Q3 (iii): A is true, but R is false.

The correct statement is (ii) Both A and R are true, and R is the correct explanation of A. Explanation: Assertion (A) is true. The bell jar experiment described in the chapter demonstrates that as air is removed from the jar (creating a vacuum), the sound of the ringing bell becomes fainter and…

Q3 (iv): A is false, but R is true.

The correct statement is (ii) Both A and R are true, and R is the correct explanation of A. Explanation: Assertion (A) is true. The bell jar experiment described in the chapter demonstrates that as air is removed from the jar (creating a vacuum), the sound of the ringing bell becomes fainter and…

Q4 (iii): A is true, but R is false.

The correct statement is (iii) A is true, but R is false. Explanation: Assertion (A) is true. A sound wave propagates through a medium as a series of compressions (regions of high density) and rarefactions (regions of low density). This disturbance travels away from the source. Reason (R) is false.…

Q4 (iv): A is false, but R is true.

The correct statement is (iii) A is true, but R is false. Explanation: Assertion (A) is true. A sound wave propagates through a medium as a series of compressions (regions of high density) and rarefactions (regions of low density). This disturbance travels away from the source. Reason (R) is false.…

Sound Waves: Characteristics and Applications Class 9 Science NCERT Solutions

Q3Pause and Ponder

Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out. Reason (R): Sound requires a medium to travel. Choose the correct statement:

(i)

Both A and R are true, but R is not the correct explanation of A.

(ii)

Both A and R are true, and R is the correct explanation of A.

(iii)

A is true, but R is false.

(iv)

A is false, but R is true.

Solution

The correct statement is (ii) Both A and R are true, and R is the correct explanation of A.

Explanation:

Assertion (A) is true. The bell jar experiment described in the chapter demonstrates that as air is removed from the jar (creating a vacuum), the sound of the ringing bell becomes fainter and eventually cannot be heard. This is a standard observation.
Reason (R) is also true. Sound waves are mechanical waves, which means they are vibrations of particles in a medium. They cannot travel through a vacuum where there are no particles to vibrate.
The reason correctly explains the assertion. We cannot hear the bell in a vacuum precisely because sound requires a material medium (like air) to propagate, and the vacuum pump has removed that medium.

Q4Pause and Ponder

Assertion (A): Compressions and rarefactions move through the medium. Reason (R): Individual particles of the medium continuously move forward with the wave. Choose the correct statement:

(i)

Both A and R are true, but R is not the correct explanation of A.

(ii)

Both A and R are true, and R is the correct explanation of A.

(iii)

A is true, but R is false.

(iv)

A is false, but R is true.

Solution

The correct statement is (iii) A is true, but R is false.

Explanation:

Assertion (A) is true. A sound wave propagates through a medium as a series of compressions (regions of high density) and rarefactions (regions of low density). This disturbance travels away from the source.
Reason (R) is false. The particles of the medium do not travel with the wave. Instead, they oscillate back and forth about their fixed mean positions. It is the energy and the disturbance that move forward, not the particles themselves.

Q5Pause and Ponder

When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?

(i)

Air particles near the tuning fork

(ii)

Energy carried by sound waves

(iii)

The tuning fork material

(iv)

A continuous stream of compressed air

Solution

The correct option is (ii) Energy carried by sound waves.

Explanation: Sound is a form of energy that propagates as a wave. The vibrating tuning fork transfers energy to the adjacent air particles, causing them to vibrate. These particles then transfer the energy to their neighbors, and this process continues until the energy reaches your ear. The air particles themselves only oscillate around their mean positions; they do not travel from the tuning fork to your ear. Therefore, it is the disturbance, or the energy of the wave, that is transmitted through the medium.

Q6Pause and Ponder

The variation of density of the medium for two sound waves is shown in Fig. 10.17 (a) and (b). Label compression and rarefaction by C and R on it. In the graph given in Fig. 10.17 (c) and (d), label the axes and draw the curves corresponding to Fig. 10.17 (a) and (b).

Solution

1. Labeling Compressions (C) and Rarefactions (R): In the diagrams of particle distribution (Fig. 10.17 (a) and (b)), the regions where the particles are crowded together are compressions (C), and the regions where the particles are spread apart are rarefactions (R).

For Fig. 10.17 (a) and (b): You should mark a 'C' over each dense cluster of dots and an 'R' over each sparse region of dots.

2. Labeling Axes and Drawing Curves: In the graphs (Fig. 10.17 (c) and (d)), we represent the sound waves by plotting density against distance.

Axes Labeling:
- The horizontal axis (x-axis) should be labeled Distance.
- The vertical axis (y-axis) should be labeled Density.
- A horizontal dashed line should be drawn across the middle to represent the Average Density.
Drawing the Curves:
- Compression (C) corresponds to a peak (crest) on the graph, where the density is maximum (above average).
- ** rarefaction (R)** corresponds to a valley (trough) on the graph, where the density is minimum (below average).
- Fig. 10.17 (a) shows more oscillations in the same length compared to (b). This means it has a shorter wavelength. The corresponding graph in (c) should show a wave with more cycles (shorter wavelength).
- Fig. 10.17 (b) shows fewer oscillations, meaning a longer wavelength. The corresponding graph in (d) should show a wave with fewer cycles (longer wavelength).

The drawn curves should be smooth, wave-like shapes oscillating above and below the average density line, with peaks aligning with compressions and troughs aligning with rarefactions from the particle diagrams.

Q7Pause and Ponder

Conduct Activity 10.1 once again with a thick rubber band and then with a thin rubber band. Does the thin rubber band vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber band?

Solution

Yes, when stretched to a similar tension, the thin rubber band will vibrate faster than the thick rubber band.

This is because the thin rubber band has less mass per unit length. Lighter objects can change their direction of motion more quickly, leading to a faster rate of vibration.

Difference in Frequency and Time Period:

Frequency: Since the thin rubber band vibrates faster, it completes more vibrations in a given amount of time. Therefore, the sound produced by the thin rubber band has a higher frequency than the sound from the thick rubber band. This would be perceived as a higher-pitched sound.
Time Period: The time period ( $T$ ) is the time taken for one complete vibration. It is the inverse of frequency ( $T = 1/\nu$ ). Because the thin rubber band has a higher frequency, it will have a shorter time period compared to the thick rubber band.

Q8Pause and Ponder

If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?

Solution

Given: Frequency of the piston,

\nu = 20 \text{ Hz}

Time,

t = 1 \text{ minute}

To Find: Number of oscillations in one minute.

Concept: Frequency is defined as the number of oscillations per second. A frequency of 20 Hz means the piston completes 20 oscillations in 1 second.

Calculation: First, convert the time from minutes to seconds.

t = 1 \text{ minute} = 60 \text{ seconds}

Now, calculate the total number of oscillations:

\text{Number of oscillations} = \text{Frequency} \times \text{Time in seconds}

\text{Number of oscillations} = 20 \text{ oscillations/second} \times 60 \text{ seconds}

\text{Number of oscillations} = 1200

Final Answer: The piston completes 1200 oscillations per minute.

Q9Pause and Ponder

For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?

Solution

From the Graph (Fig. 10.19): The graph shows the variation of density with distance. The wavelength (

\lambda

) is the distance of one complete wave cycle.

Identify the wavelength: Looking at the x-axis (Distance in cm), the wave starts at 0, completes a full cycle, and returns to the axis at 40 cm. Therefore, the wavelength of the sound wave is 40 cm. $\lambda = 40 \text{ cm}$
Calculate half of the wavelength: $\text{Half wavelength} = \frac{\lambda}{2}$ $\text{Half wavelength} = \frac{40 \text{ cm}}{2} = 20 \text{ cm}$

Final Answer: Half of the wavelength of the sound wave is 20 cm.

Q10Pause and Ponder

Table 10.1 shows the speed of sound in a few media at atmospheric pressure.
State Substance/Medium Approximate speed
Solid Steel $5000 \text{ m s}^{-1}$
Liquid Water $1500 \text{ m s}^{-1}$
Gas Air $340 \text{ m s}^{-1}$
Compare the speeds in different media by finding the ratio of

State	Substance/Medium	Approximate speed
Solid	Steel	$5000 \text{ m s}^{-1}$
Liquid	Water	$1500 \text{ m s}^{-1}$
Gas	Air	$340 \text{ m s}^{-1}$

(i)

the speed of sound in water with respect to the speed in the air.

(ii)

the speed of sound in steel with respect to the speed in the water.

Solution

Given Data from Table 10.1:

Speed of sound in Air ( $v_{\text{air}}$ ) = $340 \text{ m s}^{-1}$
Speed of sound in Water ( $v_{\text{water}}$ ) = $1500 \text{ m s}^{-1}$
Speed of sound in Steel ( $v_{\text{steel}}$ ) = $5000 \text{ m s}^{-1}$

(i) Ratio of the speed of sound in water to the speed in air:

Formula:

\text{Ratio} = \frac{\text{Speed in water}}{\text{Speed in air}} = \frac{v_{\text{water}}}{v_{\text{air}}}

Calculation:

\text{Ratio} = \frac{1500 \text{ m s}^{-1}}{340 \text{ m s}^{-1}} \approx 4.41

Final Answer (i): The speed of sound in water is approximately 4.41 times the speed of sound in air.

(ii) Ratio of the speed of sound in steel to the speed in water:

Formula:

\text{Ratio} = \frac{\text{Speed in steel}}{\text{Speed in water}} = \frac{v_{\text{steel}}}{v_{\text{water}}}

Calculation:

\text{Ratio} = \frac{5000 \text{ m s}^{-1}}{1500 \text{ m s}^{-1}} \approx 3.33

Final Answer (ii): The speed of sound in steel is approximately 3.33 times the speed of sound in water.

Q11Pause and Ponder

Two friends are standing along a steel fence at a distance of 340 m from each other (Fig. 10.23). Gunjan places her ear over the fence and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air given in Table 10.1, calculate the time difference between the sound that reached Gunjan through the air and the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least 0.1 s to be heard separately.)

Solution

Given:

Distance between friends, $d = 340 \text{ m}$
Speed of sound in air, $v_{\text{air}} = 340 \text{ m s}^{-1}$ (from Table 10.1)
Speed of sound in steel, $v_{\text{steel}} = 5000 \text{ m s}^{-1}$ (from Table 10.1)
Minimum time interval for distinguishing sounds, $\Delta t_{\text{min}} = 0.1 \text{ s}$

To Find:

The time difference ( $\Delta t$ ) between the arrival of sound through air and steel.
Whether the two sounds can be distinguished.

Formula: Time =

\frac{\text{Distance}}{\text{Speed}}

Calculation:

Time taken for sound to travel through air ( $t_{\text{air}}$ ): $t_{\text{air}} = \frac{d}{v_{\text{air}}} = \frac{340 \text{ m}}{340 \text{ m s}^{-1}} = 1 \text{ s}$
Time taken for sound to travel through the steel fence ( $t_{\text{steel}}$ ): $t_{\text{steel}} = \frac{d}{v_{\text{steel}}} = \frac{340 \text{ m}}{5000 \text{ m s}^{-1}} = 0.068 \text{ s}$
Time difference ( $\Delta t$ ): The sound through steel arrives first as it travels faster. $\Delta t = t_{\text{air}} - t_{\text{steel}} = 1 \text{ s} - 0.068 \text{ s} = 0.932 \text{ s}$

Conclusion: The calculated time difference is

0.932 \text{ s}

. Since

0.932 \text{ s} > 0.1 \text{ s}

, the time interval between the arrival of the two sounds is greater than the minimum time required for the human ear to distinguish them.

Final Answer: The time difference between the sound reaching Gunjan through the air and the steel is 0.932 s. Yes, it would have been possible for her to distinguish between the two sounds because the time gap is larger than 0.1 s.

Q12Pause and Ponder

An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be $343 \text{ m s}^{-1}$ .

Solution

Given:

Time interval for the echo, $t = 0.2 \text{ s}$
Speed of sound, $v = 343 \text{ m s}^{-1}$

To Find:

The minimum distance ( $d$ ) of the reflecting surface.

Concept: An echo is heard when sound travels from the source to a reflecting surface and then back to the listener. The time given (

t = 0.2 \text{ s}

) is the total time for this round trip.

Formula: Total distance traveled by sound = Speed

\times

Time

2d = v \times t

Therefore, the distance to the surface is:

d = \frac{v \times t}{2}

Calculation:

d = \frac{343 \text{ m s}^{-1} \times 0.2 \text{ s}}{2}

d = \frac{68.6 \text{ m}}{2}

d = 34.3 \text{ m}

Final Answer: The minimum distance the reflecting surface should be placed at is 34.3 m.

Q13Pause and Ponder

Sound travels much farther in water than light, and thus, is used for various underwater applications. A sonar signal sent to find the depth of ocean takes 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is $1500 \text{ m s}^{-1}$ ?

Solution

Given:

Total time for the sonar signal to return (round trip), $t = 4 \text{ s}$
Speed of sound in seawater, $v = 1500 \text{ m s}^{-1}$

To Find:

The depth of the ocean ( $d$ ).

Concept: The total time of 4 s includes the time for the signal to travel from the ship to the ocean floor and the time for it to travel back to the ship.

Formula: The one-way travel time is half of the total time. Time to reach the ocean floor,

t_{\text{one-way}} = \frac{t}{2}

The depth is the distance traveled in this one-way time.

d = v \times t_{\text{one-way}}

Calculation:

Calculate the one-way travel time: $t_{\text{one-way}} = \frac{4 \text{ s}}{2} = 2 \text{ s}$
Calculate the depth: $d = 1500 \text{ m s}^{-1} \times 2 \text{ s}$ $d = 3000 \text{ m}$

Final Answer: The depth of the ocean at that location is 3000 m, or 3 km.

Q1Revise, Reflect, Refine

Which observation best supports the idea that sound is a mechanical wave?

(i)

Sound shows reflection

(ii)

Sound needs a medium to propagate

(iii)

Sound has frequency

(iv)

Sound carries energy

Solution

The correct option is (ii) Sound needs a medium to propagate.

Explanation: A mechanical wave is defined as a wave that requires a material medium (solid, liquid, or gas) for its propagation. The fact that sound cannot travel through a vacuum, as demonstrated by the bell jar experiment, is the most direct evidence that it is a mechanical wave. While other options are true for sound, they do not uniquely define it as a mechanical wave. For instance, light (an electromagnetic wave) also shows reflection, has frequency, and carries energy, but it does not need a medium.

Q2Revise, Reflect, Refine

For a sound wave propagating in a medium, increasing its frequency will increase its

(i)

wavelength

(ii)

speed

(iii)

number of compressions per second

(iv)

time period

Solution

The correct option is (iii) number of compressions per second.

Explanation:

Frequency ( $\nu$ ) is defined as the number of oscillations (or cycles, which include one compression and one rarefaction) that pass a point per second. Therefore, increasing the frequency directly means increasing the number of compressions passing a point per second.
(i) Wavelength ( $\lambda$ ): For a given medium, the speed of sound ( $v$ ) is constant. The relationship is $v = \nu \times \lambda$ . If frequency ( $\nu$ ) increases, the wavelength ( $\lambda$ ) must decrease to keep the speed ( $v$ ) constant.
(ii) Speed ( $v$ ): The speed of sound depends on the properties of the medium (like temperature and density), not on the frequency of the wave.
(iv) Time period ( $T$ ): The time period is the inverse of frequency ( $T = 1/\nu$ ). If frequency increases, the time period decreases.

Q3Revise, Reflect, Refine

If 20 compressions pass a point in 4 seconds, the frequency is

(i)

80 Hz

(ii)

5 Hz

(iii)

10 Hz

(iv)

0.2 Hz

Solution

The correct option is (ii) 5 Hz.

Concept: Frequency is the number of complete oscillations (or compressions) passing a point per unit time.

Formula:

\text{Frequency (}\nu\text{)} = \frac{\text{Number of compressions}}{\text{Time taken}}

Calculation: Given:

Number of compressions = 20
Time taken = 4 seconds

\nu = \frac{20}{4 \text{ s}} = 5 \text{ s}^{-1} = 5 \text{ Hz}

Final Answer: The frequency is 5 Hz.

Q4Revise, Reflect, Refine

In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify your answer.

Solution

It will produce reverberation, not a distinct echo.

Justification: The human ear can distinguish between an original sound and its reflection only if the time interval between them is at least 0.1 seconds. This persistence of hearing allows the brain to perceive them as two separate sounds.

In this case, the time interval is 0.05 seconds, which is less than the required 0.1 seconds. When reflected sounds arrive this quickly, the brain does not perceive them as separate. Instead, the reflected sound overlaps with the original sound, causing the sound to persist for a short time after the source has stopped. This persistence of sound due to multiple reflections is called reverberation.

Q5Revise, Reflect, Refine

Graphs representing two sound waves are given in Fig. 10.30. If the scales on the X and Y axes of the two graphs are the same, which of the two sound waves has (i) greater wavelength, and (ii) smaller amplitude?

Solution

(i) Greater Wavelength:

Wave (a) has the greater wavelength.
Explanation: Wavelength is the length of one complete wave cycle measured along the horizontal (distance) axis. By observing the graphs, we can see that one complete cycle of wave (a) covers a longer distance on the x-axis compared to one complete cycle of wave (b). Therefore, wave (a) has a greater wavelength.

(ii) Smaller Amplitude:

Wave (a) has the smaller amplitude.
Explanation: Amplitude is the maximum displacement from the central (zero) line, measured along the vertical axis. The peaks and troughs of wave (a) are closer to the central line than the peaks and troughs of wave (b). This means wave (a) has a smaller maximum displacement and thus a smaller amplitude.

Q6Revise, Reflect, Refine

The sound waves emitted by three sources A, B and C are represented in Fig. 10.31. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.

Solution

Identification of Curves:

Frequency is the number of waves passing a point in a given time, or the number of full cycles within a given distance on a graph. A higher frequency corresponds to more waves packed into the same space (shorter wavelength).

Source A (Maximum Frequency): The wave with the maximum frequency will have the most number of cycles in the given length. In Fig. 10.31, the bottom curve has the highest number of oscillations and the shortest wavelength. Therefore, this curve represents source A.
Source C (Minimum Frequency): The wave with the minimum frequency will have the fewest number of cycles in the given length. The top curve has the lowest number of oscillations and the longest wavelength. Therefore, this curve represents source C.
Source B: By elimination, the middle curve, which has a frequency between A and C, represents source B.

To mark them on the figure:

Label the top curve as C.
Label the middle curve as B.
Label the bottom curve as A.

Q7Revise, Reflect, Refine

Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.

Solution

To draw the graph, follow these steps:

Set up the axes:
- The horizontal axis (x-axis) represents Distance (in cm).
- The vertical axis (y-axis) represents the Change in Density from the average.
Mark the amplitude:
- The density amplitude is 3 units. This means the wave's peaks (crests) will be at +3 on the y-axis, and its valleys (troughs) will be at -3.
Mark the wavelength:
- The wavelength is 4 cm. This means one complete cycle of the wave must be completed over a distance of 4 cm on the x-axis.
Draw the wave:
- Start the wave at the origin (0, 0).
- Draw the wave going up to its first peak (crest) at a distance of 1 cm (one-quarter of the wavelength). The coordinates will be (1, 3).
- Continue the curve downwards, crossing the x-axis at 2 cm (half the wavelength). The coordinates will be (2, 0).
- Draw the wave down to its first valley (trough) at a distance of 3 cm (three-quarters of the wavelength). The coordinates will be (3, -3).
- Bring the curve back up to the x-axis at 4 cm to complete one full wavelength. The coordinates will be (4, 0).

If you need to draw more cycles, simply repeat this pattern every 4 cm. The resulting graph will be a sinusoidal curve oscillating between +3 and -3 with a period of 4 cm along the distance axis.

Q8Revise, Reflect, Refine

In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?

Solution

There are two main scientific errors in this depiction:

Presence of Sound in Space: Sound is a mechanical wave, which means it requires a material medium (like air, water, or a solid) to travel. Outer space is a near-perfect vacuum, containing almost no particles. Therefore, sound waves cannot propagate in space. An explosion of a spacecraft would be completely silent to an outside observer.
Simultaneous Light and Sound: Even if there were a medium in space for the sound to travel through, sound and light travel at vastly different speeds. The speed of light in a vacuum is approximately $3 \times 10^8 \text{ m/s}$ , making it appear instantaneous over typical viewing distances. The speed of sound is much slower (e.g., $344 \text{ m/s}$ in air at room temperature). An observer at any distance from the explosion should see the flash of light long before they hear the sound. Depicting them at the same time is incorrect.

Q9Revise, Reflect, Refine

A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of $344 \text{ m s}^{-1}$ find its time period.

Solution

Given:

Wavelength, $\lambda = 3.44 \text{ m}$
Speed of the wave, $v = 344 \text{ m s}^{-1}$

To Find:

Time period, $T$

Formulas:

The relationship between speed, frequency, and wavelength is: $v = \nu \times \lambda$
The relationship between frequency and time period is: $T = \frac{1}{\nu}$

Combining these, we can write:

v = \frac{1}{T} \times \lambda

, which can be rearranged to find the time period.

T = \frac{\lambda}{v}

Calculation:

T = \frac{3.44 \text{ m}}{344 \text{ m s}^{-1}}

T = 0.01 \text{ s}

Final Answer: The time period of the sound wave is 0.01 s.

Q10Revise, Reflect, Refine

A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s . If ultrasonic wave travels at $1525 \text{ m s}^{-1}$ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?

Solution

Given:

Total time for the echo to return (round trip), $t = 5 \text{ s}$
Speed of the ultrasonic wave in seawater, $v = 1525 \text{ m s}^{-1}$

To Find:

The depth of the wreckage ( $d$ ).

Concept: The time of 5 s is the total time taken for the sonar signal to travel from the ship to the wreckage and then reflect back to the ship.

Formula: The distance to the wreckage is the distance covered in the one-way travel time. Time for one-way travel,

t_{\text{one-way}} = \frac{t}{2}

Depth,

d = v \times t_{\text{one-way}}

Calculation:

Calculate the one-way travel time: $t_{\text{one-way}} = \frac{5 \text{ s}}{2} = 2.5 \text{ s}$
Calculate the depth: $d = 1525 \text{ m s}^{-1} \times 2.5 \text{ s}$ $d = 3812.5 \text{ m}$

Final Answer: The wreckage of the sunken ship is located approximately 3812.5 m (or about 3.81 km) down in the ocean.

Q11Revise, Reflect, Refine

A vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz ) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be $345 \text{ m s}^{-1}$ .

Solution

Given:

Distance to the obstacle, $d = 1.2 \text{ m}$
Speed of the ultrasonic wave in air, $v = 345 \text{ m s}^{-1}$

To Find:

The total time ( $t$ ) taken for the wave to travel to the obstacle and come back.

Concept: The wave travels to the obstacle and then reflects back to the sensor. So, the total distance traveled by the wave is twice the distance to the obstacle.

Formula: Total distance =

2 \times d

Time =

\frac{\text{Total distance}}{\text{Speed}}

t = \frac{2d}{v}

Calculation:

Calculate the total distance traveled: $\text{Total distance} = 2 \times 1.2 \text{ m} = 2.4 \text{ m}$
Calculate the total time: $t = \frac{2.4 \text{ m}}{345 \text{ m s}^{-1}} \approx 0.0069565 \text{ s}$

Rounding to two significant figures, the time is approximately

0.0070 \text{ s}

.

Final Answer: The time taken by the ultrasonic wave to travel to the obstacle and come back is approximately 0.0070 s (or 7.0 milliseconds).

Q12Revise, Reflect, Refine

The speed of sound in air is about $331 \text{ m s}^{-1}$ at 0 °C and nearly $344 \text{ m s}^{-1}$ at 22 °C. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 °C to 0 °C? Assume that all other conditions remain unchanged.

Solution

Given:

Distance, $d = 1720 \text{ m}$
Speed of sound at 22 °C, $v_{22} = 344 \text{ m s}^{-1}$
Speed of sound at 0 °C, $v_{0} = 331 \text{ m s}^{-1}$

To Find:

The extra time ( $\Delta t$ ) taken for the sound to travel when the temperature drops from 22 °C to 0 °C.

Formula: Time =

\frac{\text{Distance}}{\text{Speed}}

Calculation:

Time taken at 22 °C ( $t_{22}$ ): $t_{22} = \frac{d}{v_{22}} = \frac{1720 \text{ m}}{344 \text{ m s}^{-1}} = 5 \text{ s}$
Time taken at 0 °C ( $t_{0}$ ): $t_{0} = \frac{d}{v_{0}} = \frac{1720 \text{ m}}{331 \text{ m s}^{-1}} \approx 5.19637 \text{ s}$
Calculate the extra time ( $\Delta t$ ): $\Delta t = t_{0} - t_{22}$ $\Delta t \approx 5.19637 \text{ s} - 5 \text{ s} \approx 0.19637 \text{ s}$

Rounding to a reasonable number of significant figures, we get 0.20 s.

Final Answer: The sound of thunder will take roughly 0.20 s extra time to travel the distance.

Q13Revise, Reflect, Refine

The variation of density of medium for a sound wave propagating with a speed of $340 \text{ m s}^{-1}$ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.

Solution

Given:

Speed of sound, $v = 340 \text{ m s}^{-1}$
The graph of density vs. distance (Fig. 10.32).

To Find:

Wavelength ( $\lambda$ )
Frequency ( $\nu$ )

Calculation:

1. Calculate the Wavelength ( $\lambda$ ):

From the graph, we can see that one complete wave cycle (from one crest to the next, or from the start to the end of one 'S' shape) finishes at a distance of 80 cm.
The wavelength is the length of one complete cycle.
Convert the wavelength from cm to m: $\lambda = 80 \text{ cm} = 0.80 \text{ m}$

2. Calculate the Frequency ( $\nu$ ):

Formula: The relationship between speed, frequency, and wavelength is $v = \nu \times \lambda$ .
We can rearrange this to solve for frequency: $\nu = \frac{v}{\lambda}$
Substitute the known values: $\nu = \frac{340 \text{ m s}^{-1}}{0.80 \text{ m}} = 425 \text{ Hz}$

Final Answer:

The wavelength of the sound wave is 0.80 m.
The frequency of the sound wave is 425 Hz.

Q14Revise, Reflect, Refine

The graphical representation of two sound waves A and B propagating at the same speed of $345 \text{ m s}^{-1}$ is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.

Solution

Given:

Speed of both waves, $v = 345 \text{ m s}^{-1}$
Graphs for wave A and wave B (Fig. 10.33).

To Find:

Wavelength of wave A ( $\lambda_A$ ) and wave B ( $\lambda_B$ ).
Frequency of wave A ( $\nu_A$ ) and wave B ( $\nu_B$ ).

Calculation for Wave A:

Wavelength of A ( $\lambda_A$ ):
- From the graph for wave A, one complete cycle finishes at a distance of 1 m on the x-axis.
- Therefore, $\lambda_A = 1 \text{ m}$ .
Frequency of A ( $\nu_A$ ):
- Formula: $\nu_A = \frac{v}{\lambda_A}$
- Calculation: $\nu_A = \frac{345 \text{ m s}^{-1}}{1 \text{ m}} = 345 \text{ Hz}$ .

Calculation for Wave B:

Wavelength of B ( $\lambda_B$ ):
- From the graph for wave B, one complete cycle finishes at a distance of 2 m on the x-axis.
- Therefore, $\lambda_B = 2 \text{ m}$ .
Frequency of B ( $\nu_B$ ):
- Formula: $\nu_B = \frac{v}{\lambda_B}$
- Calculation: $\nu_B = \frac{345 \text{ m s}^{-1}}{2 \text{ m}} = 172.5 \text{ Hz}$ .

Final Answer:

For Wave A: Wavelength = 1 m, Frequency = 345 Hz.
For Wave B: Wavelength = 2 m, Frequency = 172.5 Hz.

Q15Revise, Reflect, Refine

Two identical sound sources are placed at A and B -one in air and one submerged in water (Fig. 10.34). Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times than that of B, what is the ratio between the speeds of sound in air and water?

Solution

Given:

Source A is in air, Source B is in water.
Let the distance to the cliff be $d$ .
Let the speed of sound in air be $v_{\text{air}}$ .
Let the speed of sound in water be $v_{\text{water}}$ .
The time taken for the echo to return to A is $t_A$ .
The time taken for the echo to return to B is $t_B$ .
Relationship between the times: $t_A = 4.5 \times t_B$ .

To Find:

The ratio of the speed of sound in air to the speed of sound in water ( $\frac{v_{\text{air}}}{v_{\text{water}}}$ ).

Formula: The time taken for an echo to return is given by

t = \frac{2d}{v}

, where

2d

is the total distance (to the cliff and back).

Calculation:

Express the times in terms of speed and distance:
- For source A (in air): $t_A = \frac{2d}{v_{\text{air}}}$
- For source B (in water): $t_B = \frac{2d}{v_{\text{water}}}$
Use the given relationship between the times: $t_A = 4.5 \times t_B$
Substitute the expressions from step 1 into the relationship from step 2: $\frac{2d}{v_{\text{air}}} = 4.5 \times \left(\frac{2d}{v_{\text{water}}}\right)$
Simplify the equation:
- The term '2d' is common on both sides and can be cancelled out. $\frac{1}{v_{\text{air}}} = \frac{4.5}{v_{\text{water}}}$
Rearrange the equation to find the required ratio $\frac{v_{\text{air}}}{v_{\text{water}}}$ :
- Cross-multiply to get: $v_{\text{water}} = 4.5 \times v_{\text{air}}$
- Divide both sides by $v_{\text{water}}$ and by 4.5: $\frac{v_{\text{air}}}{v_{\text{water}}} = \frac{1}{4.5}$
Convert the decimal to a fraction: $\frac{1}{4.5} = \frac{1}{9/2} = \frac{2}{9}$

Final Answer: The ratio between the speed of sound in air and water is 1 : 4.5 or 2 : 9.

Sound Waves: Characteristics and ApplicationsClass 9 Science NCERT Solutions

Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out. Reason (R): Sound requires a medium to travel. Choose the correct statement:

Both A and R are true, but R is not the correct explanation of A.

Both A and R are true, and R is the correct explanation of A.

A is true, but R is false.

A is false, but R is true.

Solution

Assertion (A): Compressions and rarefactions move through the medium. Reason (R): Individual particles of the medium continuously move forward with the wave. Choose the correct statement:

Both A and R are true, but R is not the correct explanation of A.

Both A and R are true, and R is the correct explanation of A.

A is true, but R is false.

A is false, but R is true.

Solution

When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?

Air particles near the tuning fork

Energy carried by sound waves

The tuning fork material

A continuous stream of compressed air

Solution

The variation of density of the medium for two sound waves is shown in Fig. 10.17 (a) and (b). Label compression and rarefaction by C and R on it. In the graph given in Fig. 10.17 (c) and (d), label the axes and draw the curves corresponding to Fig. 10.17 (a) and (b).

Solution

Conduct Activity 10.1 once again with a thick rubber band and then with a thin rubber band. Does the thin rubber band vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber band?

Solution

If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?

Solution

For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?

Solution

the speed of sound in water with respect to the speed in the air.

the speed of sound in steel with respect to the speed in the water.

Solution

Solution

An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 m s−1343 \text{ m s}^{-1}343 m s−1.

Solution

Solution

Which observation best supports the idea that sound is a mechanical wave?

Sound shows reflection

Sound needs a medium to propagate

Sound has frequency

Sound carries energy

Solution

For a sound wave propagating in a medium, increasing its frequency will increase its

wavelength

speed

number of compressions per second

time period

Solution

If 20 compressions pass a point in 4 seconds, the frequency is

80 Hz

5 Hz

10 Hz

0.2 Hz

Solution

In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify your answer.

Solution

Graphs representing two sound waves are given in Fig. 10.30. If the scales on the X and Y axes of the two graphs are the same, which of the two sound waves has (i) greater wavelength, and (ii) smaller amplitude?

Solution

The sound waves emitted by three sources A, B and C are represented in Fig. 10.31. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.

Solution

Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.

Solution

In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?

Solution

A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s−1344 \text{ m s}^{-1}344 m s−1 find its time period.

Solution

A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s . If ultrasonic wave travels at 1525 m s−11525 \text{ m s}^{-1}1525 m s−1 in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?

Solution

Solution

Solution

The variation of density of medium for a sound wave propagating with a speed of 340 m s−1340 \text{ m s}^{-1}340 m s−1 is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.

Solution

The graphical representation of two sound waves A and B propagating at the same speed of 345 m s−1345 \text{ m s}^{-1}345 m s−1 is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.

Solution

Solution

An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be $343 \text{ m s}^{-1}$ .

A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of $344 \text{ m s}^{-1}$ find its time period.

A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s . If ultrasonic wave travels at $1525 \text{ m s}^{-1}$ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?

The variation of density of medium for a sound wave propagating with a speed of $340 \text{ m s}^{-1}$ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.

The graphical representation of two sound waves A and B propagating at the same speed of $345 \text{ m s}^{-1}$ is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.