Sound - Class 9 Science - Chapter 11 - Notes, NCERT Solutions & Extra Questions
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Back Questions - Sound | NCERT | Science | Class 9
What is sound and how is it produced?
Sound is a form of energy that produces a sensation of hearing in our ears. It is produced by vibrating objects. When an object vibrates, it sets the particles of the medium around it vibrating, creating sound. Various actions such as plucking, scratching, rubbing, blowing, or shaking different objects can set them vibrating and thus produce sound.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Compressions and rarefactions in air near a source of sound are produced due to the longitudinal waves generated by the vibration of the sound source. When a sound source, such as a tuning fork, vibrates, it pushes on the adjacent air particles, compressing them together. This creates regions of high pressure called compressions. When the source moves back, it creates an area of lower pressure as the air particles spread out, called rarefactions. This movement of the sound source sets up a series of compressions and rarefactions that move away from the source as a sound wave.
Here is a simple diagram depicting this process:
R C R C R C R
<---- Direction of Wave Propagation
|--λ--|
C: Compression (High pressure, air particles are close together)
R: Rarefaction (Low pressure, air particles are spread apart)
λ: Wavelength, the distance between successive compressions or rarefactions
This diagram does not capture all the details of wave propagation but provides a basic visualization of the pattern of alternating high-pressure and low-pressure regions created in the air by the source of sound. The pattern moves outward from the source in the form of longitudinal waves, which is the form of sound waves in air.
Why is sound wave called a longitudinal wave?
Sound waves are called longitudinal waves because they involve the vibration of particles of the medium in a direction parallel to the direction of the wave propagation. When sound is produced, it creates a series of compressions and rarefactions in the medium, where particles move back and forth in the same direction as the sound wave is moving. This to-and-fro motion is characteristic of longitudinal waves, distinguishing them from transverse waves, where particles vibrate perpendicular to the wave direction.
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
The characteristic of the sound that helps you identify your friend by his voice in a dark room is the quality or timbre of the sound. This distinctive quality enables us to distinguish one sound from another having the same pitch and loudness. Timbre defines the richness and the unique aspects of the sound, making it possible to recognize different voices. Even in the absence of light, it's the unique timbre of your friend's voice that allows you to identify them among others.
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Flash and thunder are produced simultaneously, but they reach the observer at different times because light travels much faster than sound. The speed of light in the air is approximately 300,000 kilometers per second, allowing the flash to be seen almost instantly. In contrast, the speed of sound is around 344 meters per second at 22°C, causing a noticeable delay in hearing thunder. This difference in velocities means thunder is heard several seconds after the flash is seen.
A person has a hearing range from $20 \mathrm{~Hz}$ to $20 \mathrm{kHz}$. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as $344 \mathrm{~m} \mathrm{~s}^{-1}$.
The wavelength of a sound wave in air can be calculated using the formula: $$ \lambda = \frac{v}{f} $$ where \ ( \lambda ) is the wavelength of the sound wave, \ ( v ) is the speed of sound in air, and \ ( f ) is the frequency of the sound wave.
Given the speed of sound in air as ( 344 , \text{m/s} ), we can calculate the wavelengths corresponding to the two frequencies at the extremes of the human hearing range:
-
For the low end of the range at ( 20 , \text{Hz} ) (low-pitched sound): $$ \lambda_{low} = \frac{344 , \text{m/s}}{20 , \text{Hz}} $$
-
For the high end of the range at ( 20 , \text{kHz} ) (high-pitched sound): $$ \lambda_{high} = \frac{344 , \text{m/s}}{20 \times 10^3 , \text{Hz}} $$
Let's calculate these two values. It seems there was an issue with processing the query. Let's attempt to calculate the wavelengths using the formula directly.
For the low frequency (20 Hz): $$ \lambda_{low} = \frac{344 , \text{m/s}}{20 , \text{Hz}} = \frac{344}{20} , \text{m} $$ $$ \lambda_{low} = 17.2 , \text{m} $$
For the high frequency (20 kHz): $$ \lambda_{high} = \frac{344 , \text{m/s}}{20 \times 10^3 , \text{Hz}} = \frac{344}{20 \times 10^3} , \text{m} $$ $$ \lambda_{high} = 0.0172 , \text{m} $$
So, the typical wavelength of a sound wave at 20 Hz is approximately 17.2 meters, and the wavelength at 20 kHz is approximately 0.0172 meters (or 17.2 millimeters).
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
The speed of sound in air is approximately ( 340 , \text{m/s} ) and the speed of sound in aluminium is about ( 5100 , \text{m/s} ).
To find the ratio of times taken by the sound wave in air and aluminium to reach the second child, we make use of the relationship:
$$ \text{Time} = \frac{\text{Distance}}{\text{Speed}} $$
Assuming the distance between the two children is the same for both cases of travelling through air and aluminium, the ratio of times taken is inversely proportional to the ratio of speeds of sound in the two media. Therefore, the ratio of times is given by:
$$ \text{Ratio of times} = \frac{\text{Time in air}}{\text{Time in aluminium}} = \frac{\text{Speed in aluminium}}{\text{Speed in air}} $$
Hence, the ratio of times is:
$$ \text{Ratio of times} = \frac{5100}{340} = 15 $$
So, the sound wave in aluminium takes 1/15 of the time it takes in air to reach the second child.
The frequency of a source of sound is $100 \mathrm{~Hz}$. How many times does it vibrate in a minute?
The frequency of a source of sound is given as (100 \ \text{Hz}), which means it vibrates 100 times per second. To find out how many times it vibrates in a minute, we can multiply the frequency by the number of seconds in a minute:
$$ \text{Number of vibrations in a minute} = \text{Frequency (vibrations per second)} \times \text{Seconds in a minute} $$
$$ \text{Number of vibrations in a minute} = 100 \ \text{Hz} \times 60 \ \text{seconds} $$
$$ \text{Number of vibrations in a minute} = 6000 \ \text{vibrations} $$
So the source of sound vibrates 6000 times in a minute.
Does sound follow the same laws of reflection as light does? Explain.
Yes, sound follows the same laws of reflection as light. Similar to light, when sound reflects at the surface of a solid or liquid, the directions of the incident and reflected sound make equal angles with the normal to the reflecting surface at the point of incidence, and all three lie in the same plane. This behavior aligns with the laws of reflection studied for light, demonstrating that sound and light share fundamental reflective properties, despite differing in their physical nature.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
When a sound is reflected from a distant object, an echo is produced. For a distinct echo to be heard, the minimum distance of the obstacle from the source must be half of the distance sound travels in 0.1 s. The speed of sound increases with temperature. Thus, on a hotter day, since sound travels faster, it covers more distance in the same time. Consequently, if the distance between the reflecting surface and the source remains the same, an echo can still be heard on a hotter day.
Give two practical applications of reflection of sound waves.
Two practical applications of reflection of sound waves include:
1. Megaphones, loudhailers, and horns: These devices are designed to direct sound towards a specific direction without spreading it all around. They utilize the principle of reflection to guide sound waves forward towards the audience.
2. Stethoscopes: Used by medical professionals, stethoscopes aid in listening to sounds produced within the body, such as the heart or lungs. They employ multiple reflections of sound to convey the sound of the patient's heartbeat or respiratory sounds to the doctor's ears.
A stone is dropped from the top of a tower $500 \mathrm{~m}$ high into a pond of water at the base of the tower. When is the splash heard at the top? Given, $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ and speed of sound $=$ $340 \mathrm{~m} \mathrm{~s}^{-1}$.
To determine when the splash is heard at the top of the tower, we have to calculate two things:
- The time it takes for the stone to fall from the top of the tower into the pond.
- The time it takes for the sound of the splash to travel back up to the top of the tower.
Let's start by calculating the time it takes for the stone to fall, using the kinematic equation for an object under constant acceleration:
$$ d = \frac{1}{2} g t^2 $$
Where:
- ( d ) is the distance the stone falls, which is ( 500 , \text{m} ),
- ( g ) is the acceleration due to gravity, which is ( 10 , \text{m/s}^2 ),
- ( t ) is the time it takes for the stone to fall.
Solving for ( t ), the time taken by the stone to reach the pond:
$$ t^2 = \frac{2d}{g} $$
$$ t = \sqrt{\frac{2d}{g}} $$
Let's plug in the values:
$$ t = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10 , \text{seconds} $$
Now let's calculate the time it takes for the sound to travel back up. The speed of sound is ( 340 , \text{m/s} ). The time ( t_s ) taken by the sound to travel a distance ( d ) is given by the equation ( t_s = \frac{d}{v} ), where ( v ) is the speed of sound.
$$ t_s = \frac{500}{340} \approx 1.47 , \text{seconds} $$
So, the total time ( T ) when the splash is heard at the top is the sum of the time the stone takes to fall and the time the sound takes to travel back up:
$$ T = t + t_s = 10 + 1.47 \approx 11.47 , \text{seconds} $$
Therefore, the splash is heard approximately ( 11.47 , \text{seconds} ) after the stone is dropped.
A sound wave travels at a speed of $339 \mathrm{~m} \mathrm{~s}^{-1}$. If its wavelength is $1.5 \mathrm{~cm}$, what is the frequency of the wave? Will it be audible?
The frequency of a wave is given by the speed of the wave divided by its wavelength. The formula for calculating the frequency (( f )) of a wave is:
$$ f = \frac{v}{\lambda} $$
where:
- ( v ) is the speed of the wave,
- ( \lambda ) is the wavelength.
Given:
- ( v = 339 , \text{m/s} ),
- ( \lambda = 1.5 , \text{cm} ) (which needs to be converted to meters to match the units of speed; ( 1 , \text{cm} = 0.01 , \text{m} )).
First, let's convert the wavelength to meters:
$$ \lambda = 1.5 , \text{cm} \times 0.01 , \text{m/cm} = 0.015 , \text{m} $$
Now we can calculate the frequency:
$$ f = \frac{339 , \text{m/s}}{0.015 , \text{m}} = 22600 , \text{Hz} $$
The average audible range for humans is typically cited as being from about 20 Hz to 20,000 Hz (or 20 kHz). Since the calculated frequency, ( f = 22600 , \text{Hz} ) (or 22.6 kHz), is above this range, the sound of this wave would not be audible to most people.
What is reverberation? How can it be reduced
Reverberation is the persistence of sound in a venue due to repeated reflections off surfaces until the sound is no longer audible. Excessive reverberation is undesired as it can muddle sound clarity. To reduce reverberation, surfaces like the roof and walls of an auditorium are covered with sound-absorbent materials such as compressed fiberboard, rough plaster, or draperies.
What is loudness of sound? What factors does it depend on?
Loudness of sound is a measure of the response of the ear to the sound. It is determined primarily by the amplitude of the sound wave; the larger the amplitude, the louder the sound. Loudness is a subjective measure and also depends on the sensitivity of the listener's ear. Although it is related to the intensity of sound, which is the amount of sound energy passing each second through a unit area, loudness and intensity are not the same due to the ear's varying sensitivity to different frequencies.
How is ultrasound used for cleaning?
Ultrasound is extensively utilized for cleaning objects in hard-to-reach places, like spiral tubes or electronic components. During this process, objects are placed in a cleaning solution and ultrasonic waves are directed into it. The high frequency of these waves causes dust, grease, and dirt particles to detach from the objects. Consequently, the objects are thoroughly cleaned as the ultrasonic waves ensure that even the most challenging contaminants are removed effectively, showcasing the power of ultrasound in achieving deep cleaning.
Explain how defects in a metal block can be detected using ultrasound.
Ultrasound waves can detect defects in metal blocks, a crucial process in ensuring the integrity of materials in construction and manufacturing. The method involves sending ultrasonic waves through the metal block. Where there's a defect, such as a crack or hole, the ultrasound is reflected back. This reflection is detected by using detectors. Unlike ordinary sound, ultrasound's high frequency allows it to identify even small defects without being diffracted around them. This technique is essential for maintaining the strength and safety of structures and equipment.
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Ask Chatterbot AIExtra Questions - Sound | NCERT | Science | Class 9
The speed of sound in air is $340 \mathrm{~m}$ per second. A boy hears a bullet hitting a wall $\frac{11}{5}$ seconds after he sees the bullet collide with it. How far is the boy from the wall?
A) $737 \mathrm{~m}$
B) $816 \mathrm{~m}$
C) $759 \mathrm{~m}$
D) $748 \mathrm{~m}$
E) None of these
The correct option is D, representing a distance of 748 m.
Given:
The speed of sound in air is $340 , \mathrm{m/s}$.
The time delay between seeing and hearing the bullet's impact is $\frac{11}{5}$ seconds.
To find the distance from the boy to the wall, we use the formula for distance: $$ \text{Distance} = \text{Speed} \times \text{Time} $$ Substituting the known values: $$ \text{Distance} = 340 , \mathrm{m/s} \times \frac{11}{5} , \mathrm{s} = 340 \times 2.2 = 748 , \mathrm{m} $$
Thus, the boy is 748 meters away from the wall.
The given graph shows the displacement versus time relation for a disturbance traveling with a velocity of $1500 \mathrm{~m} \mathrm{~s}^{-1}$. Calculate the wavelength of the disturbance.
A) $3 \times 10^{-5} \mathrm{~m}$ B) $4 \times 10^{-5} \mathrm{~m}$ C) $8 \times 10^{-5} \mathrm{~m}$ D) $5 \times 10^{-5} \mathrm{~m}$
To determine the wavelength of the disturbance indicated in the graph, we use the given velocity and time period.
Given:
Velocity, $ v = 1500 , \text{m/s} $
Time period, $ T = 2 \times 10^{-6} , \text{s} $
Using the relationship between wavelength $\lambda$, velocity $v$, and time period $T$ (since frequency $ f = \frac{1}{T} $):
$$ \lambda = v \cdot T $$
Substituting the values:
$$ \lambda = 1500 , \text{m/s} \cdot 2 \times 10^{-6} , \text{s} $$
$$ \lambda = 3 \times 10^{-3} , \text{m} $$
Thus, the wavelength of the disturbance is $3 \times 10^{-3} , \text{m}$, corresponding to option A.
A bat emitting an ultrasonic wave of frequency $4.5 \times 10^{4} \mathrm{~Hz}$ flies at a speed of $6 \mathrm{~m} / \mathrm{s}$ between two parallel walls. Find the two frequencies heard by the bat and the beat frequency between the two. The speed of sound is $330 \mathrm{~m} / \mathrm{s}$.
Let us designate the walls as $\mathrm{W}{1}$ and $\mathrm{W}{2}$, with the bat flying between them. The bat emits an ultrasonic wave with a frequency $f = 4.5 \times 10^4 , \mathrm{Hz}$ and is moving at a speed of $6 , \mathrm{m/s}$. The speed of sound in air is $330 , \mathrm{m/s}$. The bat will hear different frequencies as it both emits and receives the echoes bouncing off the walls due to the Doppler Effect.
Calculations for Wall $\mathrm{W}_{2}$ (bat approaching wall):
The frequency perceived by the wall $W_2$ (bat moving towards the wall): $$ f' = \frac{v_{sound} + v_{bat}}{v_{sound} - v_{listener}} \times f = \frac{330 + 6}{330} \times 4.5 \times 10^4 = \frac{336}{330} \times 4.5 \times 10^4 $$ Since the wall is stationary, $v_{listener}$ is 0 m/s in this calculation.
Now, considering that $f'$ reflects back to the approaching bat: $$ f_{B2} = \frac{v_{sound} + v_{bat}}{v_{sound}} \times f' = \frac{336}{324} \times \left(\frac{336}{330} \times 4.5 \times 10^4\right) $$
Calculations for Wall $\mathrm{W}_{1}$ (bat moving away from wall):
The frequency perceived by wall $W_1$ (bat moving away from the wall): $$ f'' = \frac{v_{sound}}{v_{sound} + v_{bat}} \times f = \frac{330}{330 + 6} \times 4.5 \times 10^{4} = \frac{330}{336} \times 4.5 \times 10^4 $$
Now, considering that $f''$ reflects back to the departing bat: $$ f_{B1} = \frac{v_{sound} - v_{bat}}{v_{sound}} \times f'' = \frac{324}{330} \times \left(\frac{330}{336} \times 4.5 \times 10^4\right) $$
Beat Frequency Calculation:
The beat frequency $f_{beat}$ is the absolute difference between the two frequencies heard by the bat: $$ f_{beat} = | f_{B2} - f_{B1} | \approx | 4.670 \times 10^{4} , \mathrm{Hz} - 4.343 \times 10^{4} , \mathrm{Hz} | = 3270 , \mathrm{Hz} $$
Thus, the two frequencies heard by the bat are approximately $4.670 \times 10^{4} , \mathrm{Hz}$ and $4.343 \times 10^{4} , \mathrm{Hz}$, and the beat frequency between them is 3270 Hz.
List out a few applications of a microphone in our day-to-day life:
A) In telephones
B) In tape recorders
C) In AC generators
D) In hearing aids
The correct applications of a microphone in everyday life are:
A) In telephones
B) In tape recorders
D) In hearing aids
A microphone is a device that converts sound waves into an electrical signal. This conversion is based on the principle of electromagnetic induction, where sound vibrations are transformed into electrical signals. Microphones find widespread use in various devices including telephones, tape recorders, hearing aids, as well as in radio and television broadcasting.
Below is an example of how a microphone functions, illustrating the conversion of sound waves to electrical signals:
The speed of sound is 332 metres per second. A gun is fired. Describe the locus of all the people on the earth's surface who hear the sound exactly after one second.
The locus of all the people on the Earth's surface who hear the sound exactly after one second is the circumference of a circle with a radius equal to the speed of sound multiplied by one second. The speed of sound is given as $332$ metres per second, so the radius of the circle is: $$ 332 \text{ m} \times 1 \text{ s} = 332 \text{ m} $$ Thus, the circle has a center at the location where the gun is fired, and the radius is $332$ meters. This circle defines the boundary within which all points are reachable by the sound of the gun in exactly one second.
A vibrating body is known to generate:
A) Sound energy
B) Electric energy
C) Light energy
D) Chemical energy
The correct answer is A) Sound energy.
Sound energy is produced due to the vibration of matter. Specifically, it originates when a body vibrates, causing longitudinal waves that propagate through a medium such as air, water, or solids. This transmission of vibrational energy is what we perceive as sound.
Answer:
A man is standing at a distance of $12 \mathrm{~m}$ from a cliff. Will he be able to hear a clear echo of his sound? Give a reason for your answer.
No, the person will not be able to hear an echo of his sound. This is because the minimum distance required between the source of the sound and the reflecting surface (such as a cliff) should be 17m for an echo to be heard clearly.
Explanation: Sound needs a minimum time gap of ( \frac{1}{10} ) of a second (or 0.1 seconds) to be distinguished from its echo. The speed of sound in air is approximately 344 meters per second. Therefore, the minimum distance sound needs to travel to create an audible echo is calculated as follows: $$ \text{Distance} = \text{Speed} \times \text{Time} = 344 , \text{m/s} \times 0.1 , \text{s} = 34.4 , \text{m} $$ Since this distance must cover the journey to the reflecting surface and back, the single distance to the reflecting surface should be half of 34.4 m: $$ \frac{34.4 , \text{m}}{2} = 17.2 , \text{m} $$ As the man is standing only 12 m from the cliff, which is less than the required 17.2 m, he will not hear a clear echo.
(1) While you were on the Moon, would you be able to talk to each other? If not, then what is the alternative? (2) What is a sonic boom? How is it caused? What are its effects?
Solution
(a) Communication on the Moon
Sound waves require a material medium such as solids, liquids, or gases to travel. However, the Moon lacks an atmosphere to carry sound waves, which means we cannot talk on the Moon as we do on Earth.
To communicate, astronauts must use radio transmissions. This technology allows them to converse despite the vacuum environment of space.
(b) Sonic Boom: Explanation and Effects
A sonic boom results from shock waves formed when an object moves through the air at speeds exceeding the speed of sound. This phenomenon resembles a loud explosion and can be startling. The process occurs as follows:
- As an object travels faster than sound, it compresses air molecules in front of it while leaving behind a trail of lower pressure.
- This rapid compression and decompression create strong shock waves.
- These shock waves generate substantial sound energy, manifesting as the distinct and intense noise known as a sonic boom.
The effects of a sonic boom can be significant, ranging from shattering glass to potentially causing damage to buildings. This impact is due to the high energy and rapid pressure changes associated with the shock waves.
An oil tanker explodes in the sea. At that instant, an aircraft and a submarine were present vertically above and below the oil tanker, respectively, at equal distances from it. Find the ratio of the time taken by sound waves to reach them. (Speed of sound in air $=340 \mathrm{~m/s}$. Speed of sound in seawater $=1520 \mathrm{~m/s}$)
A) 5.40 B) 4.47 C) 5.57 D) 6.40
The correct answer is Option B: 4.47.
Let's assume the distance from the point of explosion to both the aircraft above and the submarine below is $x$ meters.
The time taken by sound to reach the aircraft is given by: $$ t_{\text{air}} = \frac{x}{340} $$ Here, $340 , \text{m/s}$ is the speed of sound in air.
The time taken by sound to reach the submarine is given by: $$ t_{\text{water}} = \frac{x}{1520} $$ Here, $1520 , \text{m/s}$ is the speed of sound in seawater.
To find the ratio of the time taken by sound to reach the aircraft to the time it takes to reach the submarine, we compute: $$ \frac{t_{\text{air}}}{t_{\text{water}}} = \frac{\left(\frac{x}{340}\right)}{\left(\frac{x}{1520}\right)} = \frac{1520}{340} = 4.47 $$
Thus, the ratio of the time taken by sound to reach the aircraft versus the submarine is 4.47.
Oxygen is 16 times heavier than hydrogen. Equal volumes of hydrogen and oxygen are mixed. The ratio of the speed of sound in the mixture to that in hydrogen is:
A) $\sqrt{\frac{2}{17}}$
B) $\sqrt{8}$
C) $\sqrt{\frac{32}{17}}$
D) $\sqrt{\frac{1}{8}}$
The correct answer is Option A: $\sqrt{\frac{2}{17}}$.
To find the ratio of the speed of sound in a mixture of hydrogen and oxygen to that in hydrogen, we use the formula for the speed of sound in a gas mixture: $$ V_{\text{mix}} = \sqrt{\frac{r R T (m+n)}{m M_1 + n M_2}} $$ where $r$ is the universal gas constant, $R$ is the specific gas constant, $T$ is temperature in Kelvin, $m$ and $n$ are the moles of hydrogen and oxygen respectively, $M_1$ and $M_2$ are the molar masses of hydrogen and oxygen.
Since the molar masses of hydrogen (H₂) and oxygen (O₂) are approximately 2 and 32 grams/mole respectively, and oxygen is given as 16 times heavier than hydrogen, we have the molar mass of oxygen as $16 \times$ molar mass of hydrogen. Also, equal volumes of hydrogen and oxygen mean equal moles in the given condition (since both are ideal gases), so $m = n = 1$, and the formula simplifies to: $$ V_{\text{mix}} = \sqrt{\frac{r R T (1+1)}{1 \times 2 + 1 \times 32}} = \sqrt{\frac{2r R T}{34}} = \sqrt{\frac{2}{17}} V_{H_2} $$ This is because the speed of sound in hydrogen $V_{H_2}$ under ideal gas equations and at the same conditions would normally be proportional to $\sqrt{\frac{r R T}{M_1}}$.
Thus, the ratio of the speed of sound in the mixture to that in hydrogen alone is: $$ \frac{V_{\text{mix}}}{V_{H_2}} = \sqrt{\frac{2}{17}} $$ This shows a decrease in the speed of sound due to the increased average molar mass of the mixture compared to hydrogen.
Velocity of sound in air is $320 \mathrm{~m/s}$. A pipe closed at one end has a length of $1 \mathrm{~m}$, neglecting end corrections. Find the frequency(s) for which the air column in the pipe can resonate.
A) $80 \mathrm{~Hz}$
B) $240 \mathrm{~Hz}$
C) $500 \mathrm{~Hz}$
D) $400 \mathrm{~Hz}$
Here's a refined solution to the question:
Solution
The correct frequency options enabling resonance in the pipe are: A) $80 \mathrm{~Hz}$, B) $240 \mathrm{~Hz}$, and D) $400 \mathrm{~Hz}$.
Given:
- Length of the pipe, $L = 1 \mathrm{~m}$
- Speed of sound in air, $v = 320 \mathrm{~m/s}$
The formula to determine the frequencies at which a closed pipe resonates is: $$ f_n = \frac{(2n - 1)v}{4L} $$ where $n$ represents the mode number. Note that for a pipe closed at one end, only odd harmonics (odd values of $2n - 1$) are produced.
Calculations for different modes are as follows:
-
For $n=1$ (first harmonic): $$ f_1 = \frac{(2 \times 1 - 1) \times 320}{4 \times 1} = \frac{320}{4} = 80 \mathrm{~Hz} $$
-
For $n=2$ (third harmonic): $$ f_2 = \frac{(2 \times 2 - 1) \times 320}{4 \times 1} = \frac{3 \times 320}{4} = 240 \mathrm{~Hz} $$
-
For $n=3$ (fifth harmonic): $$ f_3 = \frac{(2 \times 3 - 1) \times 320}{4 \times 1} = \frac{5 \times 320}{4} = 400 \mathrm{~Hz} $$
-
For $n=4$ (seventh harmonic): $$ f_4 = \frac{(2 \times 4 - 1) \times 320}{4 \times 1} = \frac{7 \times 320}{4} = 560 \mathrm{~Hz} $$
Among the given options, options A), B) and D) are the correctly matching modes of vibrations for a closed organ pipe, thus they represent possible resonance frequencies for the air column in the pipe.
A stone is dropped into a well 44 m deep, and the sound of the splash is heard after 3.12 seconds. Find the speed of sound in air (g = 9.8 m per sec).
To find the speed of sound in air, we analyze the event in two parts: the stone falling into the well, and the sound traveling back up the well.
-
Falling Time Calculation: The well is 44 m deep. Initially, the stone is at rest, so initial velocity $u = 0$. Let $t$ denote the time in seconds it takes for the stone to hit the water. Using the formula for distance covered by a freely falling object: $$ s = ut + \frac{1}{2}gt^2 $$
Substituting the given and known values: $$ 44 = 0 \times t + 0.5 \times 9.8 \times t^2 $$ This simplifies to: $$ 44 = 4.9 t^2 $$ Solving for $t$ gives: $$ t^2 = \frac{44}{4.9} \approx 8.98 \quad \text{(not exact form 8.97 as calculated)} $$ $$ t = \sqrt{8.98} \approx 3.00, \text{s} $$ This implies the stone reaches the water surface in about 3.00 seconds.
-
Sound Travel Time Calculation: After hitting the water, the splash sound needs to travel back up to the top of the well. The total time from the stone's release to hearing the splash is 3.12 seconds. Therefore, the sound traveling time is: $$ \text{Sound time} = 3.12, \text{s} - 3.00, \text{s} = 0.12, \text{s} $$
-
Speed of Sound Calculation: Now, knowing that the sound travels the same distance (44 m) in about 0.12 seconds, the speed of sound $v$ is: $$ v = \frac{\text{Distance}}{\text{Time}} = \frac{44, \text{m}}{0.12, \text{s}} \approx 366.67, \text{m/s} $$
Thus, the speed of sound in air is approximately 366.67 m/s. Note: This calculation uses an approximate value for the time sound took to travel which might vary slightly based on actual fall time rounding.
A boy is standing on top of a tower of height 85 m and throws a ball in the vertically upward direction with a certain speed. If 5.25 seconds later he hears the ball hitting the ground, then the speed with which the boy threw the ball is (take $g = 10 , \text{m/s}^2$, speed of sound in air = 340 m/s):
A) 6 m/s
B) 8 m/s
C) 10 m/s
D) 12 m/s
To solve for the initial speed ($u$) at which the boy threw the ball vertically upward, we first need to understand how the total elapsed time of 5.25 seconds breaks down into the time taken for the ball to reach the ground and the sound of the ball hitting the ground to reach the boy.
Step 1: Calculate Time Taken by Sound to Travel
The sound travels from the ground (where the ball hits) to the top of the tower where the boy is standing. The height of the tower is 85 m and the speed of sound in air is given as 340 m/s. The time ($t_{\text{sound}}$) taken by the sound to travel this distance can be calculated using the formula: $$ t_{\text{sound}} = \frac{\text{Distance}}{\text{Speed of sound}} = \frac{85}{340} = 0.25 , \text{s} $$
Step 2: Calculate Time Taken by the Ball to Travel
To find the total time taken by the ball from the moment it was thrown to the moment it hits the ground, we subtract the time taken by the sound from the total time given: $$ t_{\text{ball}} = 5.25 , \text{s} - 0.25 , \text{s} = 5 , \text{s} $$
Step 3: Use Kinematic Equation
We use the kinematic equation for the motion under gravity ($g = 10 , \text{m/s}^2$), since the vertical motion involves a constant acceleration due to gravity: $$ y = ut + \frac{1}{2}at^2 $$ where:
- $y$ = -85 m (displacement from the top of the tower to the ground, taken as negative because it is downwards),
- $u$ = initial velocity which we need to find,
- $t$ = 5 s,
- $a$ = -10 m/s² (acceleration due to gravity is downwards).
Plugging in the values: $$ -85 = u \times 5 - \frac{1}{2} \times 10 \times 5^2 $$ Simplify the equation: $$ -85 = 5u - \frac{1}{2} \times 10 \times 25 $$ $$ -85 = 5u - 125 $$ $$ 5u = -85 + 125 = 40 $$ $$ u = \frac{40}{5} = 8 , \text{m/s} $$
Answer and Conclusion
The initial speed with which the boy threw the ball is 8 m/s. Thus, the correct answer is Option B: 8 m/s.
A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms\textsuperscript{-1}. The velocity of sound in air is 300 ms\textsuperscript{-1}. If the person can hear frequencies up to 10000 Hz, the maximum value of v up to which he can hear the whistle is
A) 30√2 ms\textsuperscript{-1}
B) 15 ms\textsuperscript{-1}
C) 30 ms\textsuperscript{-1}
D) 15/√2 ms\textsuperscript{-1}
The correct choice is B) 15 ms^{-1}.
The apparent frequency ($n'$) heard by a listener, when considering the source moving and other factors like wind speed, is given by the formula: $$ n' = n \frac{v + w + v_L}{v + w - v_S} $$ Here, $n$ is the actual frequency of sound, $v$ is the velocity of sound in air, $w$ is the velocity of wind, $v_L$ is the velocity of the listener, and $v_S$ is the velocity of the source. In scenarios where the wind speed ($w$) is zero and the listener is stationary, the formula simplifies to: $$ n' = n \frac{v}{v - v_S} $$ Since the listener can only hear frequencies up to 10000 Hz, and the whistle emits sound starting from 9500 Hz, we determine the maximum source velocity ($v_S$) that allows the listener to hear the whistle using the following equation: $$ 10000 = 9500 \frac{300}{300 - v_S} $$ To find $v_S$, rearrange the equation: $$ 10000 = 9500 \times \frac{300}{300 - v_S} \ \Rightarrow 300 - v_S = 9500 \times \frac{300}{10000} \ \Rightarrow 300 - v_S = 285 \ \Rightarrow v_S = 15 \text{ ms}^{-1} $$ Therefore, the maximum value of $v_S$ for the whistle to be audible is 15 ms^{-1}.
Explain why strings of different thicknesses are provided on a stringed instrument.
Solution
Strings of varying thicknesses are used on stringed instruments primarily to achieve the generation of different frequency sound waves. This variation in frequency is crucial because it allows the instrument to produce a range of musical notes. The fundamental reason behind this phenomenon lies in the relationship between the frequency of vibration and the string's thickness.
The natural frequency of vibration of a string when it is stretched is determined by several factors including its thickness (or radius). Mathematical expression relating the frequency $(f)$ of the string's vibration to its radius $(r)$ is given by:
$$ f \propto \frac{1}{r} $$
where $r$ is the radius of the string. From this relationship, it is evident that the frequency is inversely proportional to the string's radius. Therefore, a thicker string (larger $r$) will vibrate at a lower frequency, and conversely, a thinner string (smaller $r$) will vibrate at a higher frequency. This allows for the creation of both low-pitched and high-pitched sounds on a single instrument.
The technique used to transmit audio signals in television broadcasts is
A) Amplitude Modulation
B) Frequency Modulation
C) Pulse Code Modulation
D) Time Division Multiplexing
The correct answer is B) Frequency Modulation.
Frequency Modulation (FM) is widely utilized for transmitting audio signals in television broadcasts due to its superior resistance to noise and interference compared to other methods like Amplitude Modulation (AM). This ensures clearer sound quality under various reception conditions.
The function of the voice coil in a moving coil loudspeaker is to produce vibrations.
A) True
B) False
Answer: B) False
The primary function of the voice coil in a moving coil loudspeaker is not to produce vibrations directly. Instead, it involves a different mechanism. When electrical pulses from an amplifier (which originally might be from a microphone) pass through the voice coil, the coil is situated within a magnetic field. These electrical pulses cause the voice coil to move back and forth rapidly. This motion then interacts with other components of the loudspeaker to ultimately produce sound vibrations. The direct function of the voice coil is to convert the electrical energy into mechanical motion under the influence of the magnetic field.
A train standing at a certain distance from a railway platform is blowing a whistle of frequency 500 Hz. If the speed of sound is 340 m/s, the frequency and wavelength of the sound of the whistle heard by a man running towards the engine with a speed of 10 m/s respectively are
A. 500 Hz, 0.7 m
B. 500 Hz, 0.68 m
C. 486 Hz, 0.7 m
D. 515 Hz, 0.68 m
The correct answer is Option D: 515 Hz, 0.68 m.
In this problem, we are given that the source of sound (train’s whistle) is stationary, which means the wavelength of sound does not change due to motion of the source. The wavelength $ \lambda $ can be calculated using the formula: $$ \lambda = \frac{v}{f} $$ where $ v = 340 , \text{m/s} $ (speed of sound) and $ f = 500 , \text{Hz} $ (frequency of the whistle). Plugging in the values, we get: $$ \lambda = \frac{340 , \text{m/s}}{500 , \text{Hz}} = 0.68 , \text{m} $$
The frequency of the sound heard by the man running towards the engine, $ f' $, is affected by the relative speed of the man and the sound. Since the man is moving towards the source, the effective speed of sound relative to him is $ v + \text{man's speed} = 340 + 10 = 350 , \text{m/s} $.
The modified frequency $ f' $ can be calculated as: $$ f' = \frac{\text{Speed of sound relative to the observer}}{\lambda} = \frac{350 , \text{m/s}}{0.68 , \text{m}} \approx 514.7 , \text{Hz} \approx 515 , \text{Hz} $$
Thus, the man hears the frequency as approximately 515 Hz and the wavelength remains 0.68 m. Therefore, the correct choice is Option D.
(a) What type of pollution is caused by the working of a mixer and grinder in the kitchen? (b) Why should we not put a pin or pencil in our ears?
Solution
(a) Noise Pollution
The mixer and grinder in a kitchen generate loud noise during operation, which can be a source of noise pollution for individuals in and around the house. This noise level can increase if the appliance vibrates against the hard surface of the kitchen countertop or is placed too close to a wall.
(b) Risk of Damage to the Eardrum
Inserting a pin or pencil into the ear is hazardous because these objects have pointy ends that could potentially pierce or damage the eardrum, leading to hearing impairments or other serious ear problems.
Which are all the functions of ears?
A Maintaining balance
B Smelling
C Hearing
D Tasting
The correct functions of the ears are:
- A Maintaining balance
- C Hearing
The ears play a crucial role in hearing and are composed of three parts: the outer, middle, and inner ear. The inner ear includes a fluid-filled structure known as the cochlea, which is instrumental in maintaining balance. This fluid helps regulate our equilibrium, aiding in balance when we walk or move.
If the speed of sound in air is $332 \mathrm{~m/s}$, then its speed in vacuum
A $> 332 \mathrm{~m/s}$
B $= 332 \mathrm{~m/s}$
C $< 332 \mathrm{~m/s}$
D Does not propagate in a vacuum
The correct answer is D - Does not propagate in a vacuum.
Sound waves are a type of longitudinal and mechanical wave which crucially require a medium to propagate. This is because mechanical waves transmit energy through the vibration of molecules in a medium.
Since a vacuum implies a complete absence of matter, there are no molecules to vibrate and transmit sound. Therefore, in a vacuum, the propagation of sound is impossible.
Thus, discussing the speed of sound in a vacuum is meaningless since it simply does not exist.
A source of sound $S$ has a frequency $f$. Wind is blowing from source to observer $O$ with velocity $u$. If the speed of sound with respect to air is $C$, the wavelength of sound detected by $O$ is:
A) $\frac{C+u}{f}$
B) $\frac{C-u}{f}$
C) $\frac{C(C+u)}{(C-u)f}$
D) $\frac{C}{f}$
The correct answer is Option A: $$ \frac{C+u}{f} $$
Analysis:
The net velocity of sound relative to the ground ($v'$) is the sum of the sound's velocity with respect to the medium ($v_s$) and the velocity of the medium itself ($v_m$): $$ v' = v_s + v_m $$
Here, $v_s$ represents the speed of sound relative to the air, which is $C$, and $v_m$ is the wind velocity blowing from the source to the observer, which is $u$. Thus, we get: $$ v' = C + u $$
Given that both the source and observer are stationary, the frequency of the sound remains unchanged at $f$. Let $\lambda$ denote the wavelength of the sound as detected by the observer. From the relationship between speed, frequency, and wavelength ($v' = f \times \lambda$), it follows: $$ C + u = f \lambda \Rightarrow \lambda = \frac{C+u}{f} $$
Thus, the wavelength of the sound detected by the observer is $\frac{C+u}{f}$.
The distance between a compression and its next rarefaction in a sound wave of wavelength $\lambda$ will be:
A) $\frac{\lambda}{2}$
B) $\frac{\lambda}{4}$
C) $\frac{\lambda}{8}$
D) $\frac{\lambda}{16}$
The correct answer is A:
A wavelength, denoted as $ \lambda $, represents the distance between successive compressions or rarefactions in a sound wave. Notably, the section from a compression to the subsequent rarefaction covers half this distance, which mathematically translates to: $$ \frac{\lambda}{2} $$
Thus, the distance between a compression and its next adjacent rarefaction is $\frac{\lambda}{2}$.
What determines the loudness of a sound? What is the unit of loudness? [2 MARKS]
The loudness of a sound is primarily determined by the amplitude of vibration. The greater the amplitude, the louder the sound is perceived.
The unit used to measure loudness is the decibel.
How is the sound produced in a bamboo flute?
A) By the vibrations of air inside the bamboo.
B) By the vibrations of bamboo.
C) By hitting the bamboo.
D) By a change in the direction of air.
The correct answer is A) By the vibrations of air inside the bamboo.
Sound is produced by the vibrations of the particles of the medium through which it travels. In the case of a bamboo flute, the sound is generated by the vibrations of air molecules inside the bamboo tube. When a flutist blows air into the flute, this air is set into vibrating motion, which then produces sound waves.
What is the possible audible range in which old age people could hear?
A $20 \mathrm{~Hz} - 25 \mathrm{~kHz}$
B $20 \mathrm{~Hz} - 20 \mathrm{~kHz}$
C $20 \mathrm{~Hz} - 12 \mathrm{~kHz}$
D $20 \mathrm{~Hz} - 30 \mathrm{~kHz}$
The correct answer is C $20 \mathrm{~Hz} - 12 \mathrm{~kHz}$.
Humans typically have a hearing range from $20 \mathrm{~Hz}$ to $20,000 \mathrm{~Hz}$. However, this range can vary significantly with age. Generally, older individuals experience a reduced upper limit of hearing. For elderly people, the audible frequency range often contracts to between $20 \mathrm{~Hz}$ and $12 \mathrm{~kHz}$.
A sound source produces 40 compressions and 40 rarefactions in 0.4 seconds. What is the frequency?
To determine the frequency of a sound wave, consider that:
- The sound source produces 40 compressions and 40 rarefactions in 0.4 seconds.
Given that:
- Each wave consists of one compression and one rarefaction.
Hence:
- 40 compressions and 40 rarefactions make a total of 40 waves.
The frequency of a wave (( \nu )) is calculated by dividing the number of waves by the time period over which they are observed. Thus,
$$ \nu = \frac{\text{Number of waves}}{\text{Time taken}} = \frac{40}{0.4} = 100 , \text{Hz} $$
Therefore, the sound's frequency is 100 Hz.
"If we can only hear things when they compress air and reach our ear, then how can a man standing outside a closed room hear a man's sound from inside the room?"
Sound can travel through various mediums, including solids like the walls of a room. When a person inside a room talks or makes noise, sound waves are produced, which causes the air molecules inside the room to vibrate at a certain frequency. These vibrations can transmit through the walls of the room, causing them to vibrate as well.
Even though the walls may absorb some elements of sound, especially high-frequency waves, lower frequency vibrations can still pass through them. As these vibrations reach the exterior surface of the wall, they cause the air outside to vibrate, allowing the sound to be heard by someone standing outside the room.
Thus, the ability to hear sounds from inside a closed room depends significantly on the material and thickness of the walls, as well as the frequency and intensity of the sound generated inside.
Sort the signals in ascending order of their bandwidth:
A. Video Signal
B. Speech Signal
C. TV Signal
D. Music Signal
To arrange the given signals in ascending order of their bandwidth, we first identify the bandwidth for each type of signal:
- Speech Signal: Approximately $3000 \text{ Hz}$ or $3 \text{ kHz}$
- Music Signal: $20 \text{ kHz}$
- Video Signal: $4.2 \text{ MHz}$
- TV Signal: $6 \text{ MHz}$
Now, placing these in order from the smallest to the largest we get:
- Speech Signal ($3 \text{ kHz}$)
- Music Signal ($20 \text{ kHz}$)
- Video Signal ($4.2 \text{ MHz}$)
- TV Signal ($6 \text{ MHz}$)
It's important to note that a TV signal encompasses both audio and video content, thus demanding a larger bandwidth, usually around $6 \text{ MHz}$.
Loudness of a sound is measured in:
A) hertz
B) decibel
C) metre
D) metre per second
The correct answer is B) decibel.
Decibel (dB) is the unit used to measure the loudness of sound. The amount of amplitude a sound wave has directly influences how loud it is perceived, and this is quantified in decibels.
Which of the following is/are properties of a sound wave?
A) It does not require a medium to propagate.
B) It is a longitudinal wave.
C) It is a mechanical wave.
D) Travels faster than light.
The correct answers are:
- B) It is a longitudinal wave.
- C) It is a mechanical wave.
Sound waves are typical examples of longitudinal waves, which means that the particles of the medium through which the sound is transmitted vibrate along the direction in which the sound wave moves.
Furthermore, sound waves require a medium (like air, water, or solids) to travel through, which classifies them as mechanical waves. This contradicts option A, as sound cannot travel through a vacuum where there is no medium. Additionally, sound waves travel significantly slower than light, dismissing option D.
In a big hall, the sound gets reflected inside the hall multiple times. The repeated reflection that results in this persistence of sound is called reverberation. This phenomenon can be observed in:
A) Auditoriums
B) Cinema halls
C) Indoor stadiums
D) All of the above
The correct answer is D) All of the above.
Reverberation is the phenomenon where sound gets reflected multiple times within a large space, leading to a persistence of sound even after the original sound has stopped. This effect is noticeable in various large indoor spaces such as:
- Auditoriums
- Cinema halls
- Indoor stadiums
These settings allow sound waves to bounce off walls and surfaces, creating the auditory experience known as reverberation.
Two trains A and B simultaneously start moving along parallel tracks from a station in the same direction. A starts with a constant acceleration of $2 \mathrm{~m/s}^{2}$ from rest while B starts with the same acceleration but with an initial velocity of $40 \mathrm{~m/s}$. Twenty seconds after the start, a passenger of A hears the whistle of B. If the frequency of the whistle is $1194 \mathrm{~Hz}$ and the velocity of sound in air is $322 \mathrm{~m/s}$, then the frequency observed by the passenger is:
A) $1186 \mathrm{~Hz}$
B) $1086 \mathrm{~Hz}$
C) $986 \mathrm{~Hz}$
D) $886 \mathrm{~Hz}$
The correct answer is Option B) $1086 \text{ Hz}$.
To solve this problem, we utilize key concepts from physics involving constant acceleration and the Doppler effect. The steps are as follows:
-
Calculate the Displacements: First, calculate the displacements of Train A and Train B, 20 seconds after they start accelerating.
-
Train A starts from rest with an acceleration of $2 \mathrm{~m/s}^2$. Using the equation of motion $x = ut + \frac{1}{2}at^2$, where $u$ is the initial speed (here $u = 0$ for Train A), $$ x = 0 + \frac{1}{2} \times 2 \times 20^2 = 400 , \text{m} $$
-
Train B starts with an initial speed of $40 \mathrm{~m/s}$ and, given the same acceleration, $$ x = 40 \times 20 + \frac{1}{2} \times 2 \times 20^2 = 800 + 400 = 1200 , \text{m} $$
-
-
Determine the Frequencies: Consider the Doppler effect for the frequency of the whistle as heard by the passenger in Train A. Utilize the Doppler effect formula: $$ f_{\text{app}} = f_0 \left(\frac{v + v_p}{v + v_s}\right) $$ where $f_{\text{app}}$ is the apparent frequency, $f_0$ is the frequency emitted by Train B, $v$ is the velocity of sound, $v_p$ is the velocity of the passenger (receiver), and $v_s$ is the velocity of the source (Train B).
-
Velocity of Train A ($v_p$) after 20 seconds is, $$ v_p = 0 + 2 \times 20 = 40 , \text{m/s} $$
-
Velocity of Train B ($v_s$) after 20 seconds is, $$ v_s = 40 + 2 \times 20 = 80 , \text{m/s} $$
-
-
Apply the Doppler Effect Formula: With the velocity of sound given as $322 \text{ m/s}$, and using the Doppler formula, $$ f_{\text{app}} = 1194 \left(\frac{322 + 0}{322 + 80}\right) = 1194 \times \frac{322}{402} \approx 1086 \text{ Hz} $$
Thus, the frequency observed by the passenger of Train A is approximately 1086 Hz.
An observer stands at a distance of $850 , \text{m}$ from the cliff and fires a gun. After what time will he hear the echo of gunfire if sound travels at a speed of $350 , \text{ms}^{-1}$ in the air?
A) $6.89 , \text{s}$
B) $2.42 , \text{s}$
C) $2.93 , \text{s}$
D) $4.86 , \text{s}$
Solution
The correct answer is Option D: $4.86 , \text{s}$.
The speed of sound in air is given as: $$ v = 350 , \text{ms}^{-1} $$ The distance from the observer to the cliff is: $$ d = 850 , \text{m} $$ The sound will travel to the cliff and back to the observer, thus covering a total distance of: $$ 2 \times d = 2 \times 850 , \text{m} = 1700 , \text{m} $$
Let $t$ be the time taken for the sound to travel this distance, then: $$ t = \frac{2 \times d}{v} = \frac{1700}{350} = 4.86 , \text{s} $$
Thus, Option D ($4.86 , \text{s}$) is the correct answer, as it represents the time after which the observer will hear the echo of the gunfire.
A stone is dropped into a well of 45 m depth. Sound of splash due to the stone striking the water surface in the well is heard after 3.125 s. From the given data, the velocity of sound in air is (g=10 m/s^2).
(A) 320 m/s
B 340 m/s
C 360 m/s
D None of these
The correct option is $\mathbf{C}$ 360 m/s.
To determine the velocity of sound in air, we start by calculating the time taken by the stone to reach the bottom of the well.
Step 1: Time Taken by Stone
The time 't' for the stone to reach the bottom can be found using the equation of motion:
$$ s = ut + \frac{1}{2}gt^2 $$
Given:
$s = 45 , \text{m} $
$ u = 0 , \text{m/s} $
$g = 10 , \text{m/s}^2 $
Substituting the values, we get:
$$ 45 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 ] [ 45 = 5t^2 ] [ t^2 = 9 ] [ t = \sqrt{9} = 3 , \text{s}$$
So, the time taken by the stone to reach the bottom of the well is 3 seconds.
Step 2: Time Taken by Sound
We know that the total time after the stone is dropped until the sound is heard is 3.125 seconds. Therefore, the time taken by the sound to travel back up is:
$$ 3.125 , \text{s} - 3 , \text{s} = 0.125 , \text{s} $$
Step 3: Velocity of Sound in Air
The velocity of sound in air can be calculated using the formula:
$$ v_{\text{air}} = \frac{\text{distance}}{\text{time}} $$
Here, the distance is 45 m, and the time is 0.125 s. Substituting these values:
$$ v_{\text{air}} = \frac{45 , \text{m}}{0.125 , \text{s}} $$ $$ v_{\text{air}} = 360 , \text{m/s} $$
Thus, the velocity of sound in air is 360 m/s.
A guitar was struck loudly whereas flute was relatively softer when played. Which property of the wave would be different between the two instruments?
Amplitude
Frequency
Wavelength
None of the above
The correct option is A: Amplitude
Amplitude determines the volume of sound. As a result, the amplitudes of the waves produced by the guitar and the flute are different. The guitar, being struck loudly, produces a wave with a higher amplitude, whereas the flute, being played softly, results in a wave with a lower amplitude.
Which of the following parts is found in the inner ear?
Option 1) Cochlea
Option 2) Stirrup
Option 3) Eardrum
Option 4) Iris
The correct option is Option 1: Cochlea
The ear is comprised of three main sections: the outer ear, the middle ear, and the inner ear. The cochlea is located in the inner ear. On the other hand, the stirrup (also known as the stapes) and the eardrum (tympanic membrane) are parts of the middle ear. The iris is not part of the ear at all; it is located in the eye.
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