NCERT Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications

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NCERT Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications

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Sound Waves: Characteristics and Applications

Chapter – 10

Pause And Ponder

1. Explore various ways of producing sound.

Ans: Sound is produced by the vibration of objects. It can be produced by plucking, striking, bl

owing, rubbing, shaking, or stretching vibrating objects.

2. Make a list of different types of musical instruments and identify their vibrating parts which produce sound.

Ans:

Musical Instrument	Vibrating Part
Guitar	Strings.
Violin	Strings.
Sitar	Strings.
Tabla	Stretched membrane.
Drum	Stretched membrane.
Flute	Air column inside the tube.
Trumpet	Air column inside the tube.
Harmonium	Metal reeds.
Bell	Metal body.
Cymbals	Metal plates.

Pause And Ponder

1. 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.

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

Pause And Ponder

1. 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.

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

Pause And Ponder

1. 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.

Ans: (ii) Energy carried by sound waves.

Pause And Ponder

1. 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).

Ans: (i) Label the compressions and rarefactions in Fig. 10.17 (a) and (b).

Compression (C): Region where the particles are closely packed (maximum density).

Rarefaction (R): Region where the particles are far apart (minimum density).

Fig. 10.17 (a)

From left to right:

C → R → C → R → C

Fig. 10.17 (b)

From left to right:

R → C → R → C → R

(ii) Fig. 10.17 (c): Draw the curve corresponding to Fig. 10.17 (a)

Axes:

X-axis: Distance (or Position)

Y-axis: Density of the medium

Draw a sinusoidal curve such that:

Peaks represent Compressions (C).

Troughs represent Rarefactions (R).

Density of Medium:

(iii) Fig. 10.17 (d): Draw the curve corresponding to Fig. 10.17 (b)

Axes:

X-axis: Distance (or Position).

Y-axis: Density of the medium.

Draw the same sinusoidal curve, but starting with a rarefaction.

Density of Medium:

Pause And Ponder

1. 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?

Ans: Yes. The thin rubber band vibrates faster than the thick rubber band. Therefore, the sound produced by the thin rubber band has a higher frequency and a shorter time period than the sound produced by the thick rubber band.

2. 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?

Ans: Frequency = 20 Hz = 20 oscillations per second.

Number of oscillations in 1 minute = 20 × 60.

= 1200 oscillations.

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

Ans: From the graph, the wavelength (λ) is 2 m.

Therefore,

Half of the wavelength = λ/2 = 2/2.

= 1 m.

Pause And Ponder

1. Table 10.1 shows the speed of sound in a few media at atmospheric pressure.

Table 10.1: Speed of sound in different media at 15 deg °C.

State	Substance/Medium	Approximate Speed
Solid	Steel	5000 ms^-1
Liquid	Water	1500 ms^-1
Gas	Air	340 ms^-1

Compare the speeds in different media by finding the ratio of:

(i) The speed of sound in water with respect to the speed in the air.

Ans: Speed of sound in water = 1500 m s⁻¹.

Speed of sound in air = 340 m s⁻¹.

Ratio = 1500 : 340.

= 75 : 17.

(ii) The speed of sound in steel with respect to the speed in the water.

Ans: Speed of sound in steel = 5000 m s⁻¹.

Speed of sound in water = 1500 m s⁻¹.

Ratio = 5000 : 1500.

= 10 : 3.

2. 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.)

Fig. 10.23: Time difference between the sound reaching through air and steel.

Ans: Distance = 340 m.

Time taken through air,

= Distance / Speed.

= 340 / 340.

= 1 s.

Time taken through steel,

= Distance / Speed.

= 340 / 5000.

= 0.068 s.

Time difference,

= 1 − 0.068.

= 0.932 s.

Since 0.932 s > 0.1 s, Gunjan can distinguish the two sounds because they reach her ears separately.

Pause And Ponder

1. 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.

Ans: Speed of sound = 343 m s⁻¹.

Time taken for echo = 0.2 s.

Distance travelled by sound = Speed × Time.

= 343 × 0.2.

= 68.6 m.

Since the sound travels to the reflecting surface and back,

Distance of the reflecting surface,

= 68.6 / 2.

= 34.3 m.

Therefore, the minimum distance of the reflecting surface should be 34.3 m.

Pause And Ponder

1. 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 m s¹?

Ans: Speed of sound in seawater = 1500 m s⁻¹.

Time taken for the sonar signal to return = 4 s.

Total distance travelled by the sound,

= Speed × Time.

= 1500 × 4.

= 6000 m.

Depth of the ocean,

= 6000 / 2.

= 3000 m.

Therefore, the depth of the ocean is 3000 m.

Revise, Reflect, Refine

1. 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.

Ans: (ii) Sound needs a medium to propagate.

A mechanical wave is defined as a wave that requires a material medium to propagate. Since sound cannot travel in a vacuum and requires a solid, liquid, or gas, it is a mechanical wave.

2. 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.

Ans: (iii) Number of compressions per second.

Frequency is defined as the number of density oscillations (which consist of alternating compressions and rarefactions) at a fixed point per unit time. Therefore, increasing the frequency directly increases the number of compressions passing a point per second.

3. 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

Ans: (ii) 5 Hz.

Frequency is calculated as the number of oscillations divided by the time taken. Frequency=20/4=5 Hz.

4. 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.

Ans: It will produce a reverberation. To hear a distinct echo, the time gap between the original sound and the reflected sound must be at least 0.1 s. If the time difference is less than 0.1 s (as in this case, 0.05 s), the human brain cannot distinguish them as separate sounds, leading to a persistence of sound known as reverberation.

5. 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?

Ans: (i) The wave having greater distance between two consecutive compressions (or crests) has the greater wavelength.

(ii) The wave having smaller height from the mean line has the smaller amplitude.

6. 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.

Ans: A: Green curve (maximum frequency).

B: Red curve (intermediate frequency).

C: Blue curve (minimum frequency).

Reason: Frequency is the number of complete waves (oscillations) in a given distance. The green curve has the most oscillations, the blue curve has the fewest, and the red curve lies in between.

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

Ans:

8. 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?

Ans: There are two major scientific errors in this depiction:

(i) Outer space is a near vacuum, and sound waves require a material medium to travel. Therefore, the sound of the explosion would not be heard at all.

(ii) Even if there were a medium, light travels extremely fast (almost instantaneously) compared to the much slower speed of sound. Therefore, a distant observer would see the flash of light long before hearing any sound, not at the exact same time.

9. A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s^–1 find its time period.

Ans: We know that Speed(v) = Wavelength (λ)/Time period (T).

Rearranging for time period:

T= λ/v T = 3.44 m/344 ms⁻¹

= 0.01 s.

10. 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^–1 in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?

Ans: The total time taken for the signal to travel down and bounce back is 5 s.

Time taken to reach the wreckage (one way) = 5 s/2=2.5 s

Distance=Speed×Time Distance

=1525 m s⁻¹ ×2.5

s = 3812.5 m.

11. 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 m s^–1.

Ans: The total distance the wave must travel is to the obstacle and back.

Total Distance=1.2 m (forward)+1.2 m (return)=2.4 m

Using the formula:

Time=Distance/Speed: Time=2.4 m/345 m s⁻¹

≈ 0.00696 s (or roughly 0.007 s)

12. The speed of sound in air is about 331 m s^–1 at 0 ºC and nearly 344 m s^–1 at 22 ºC. 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 ºC to 0 ºC? Assume that all other conditions remain unchanged.

Ans: Time taken at 22 ºC: t 1 = Distance/Speed

= 1720 m/344 m s⁻¹

= 5.0 s

Time taken at 0 ºC: t 2 = Distance/Speed

=1720 m/331 m s⁻¹

≈ 5.196 s

Extra time taken = 5.196 s−5.0 s

≈ 0.196 s.

13. The variation of density of medium for a sound wave propagating with a speed of 340 m s^–1 is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.

Ans: Wavelength (λ) = 8 cm=0.08 m

Frequency (ν)=Speed (v)/Wavelength (λ).

ν=340 m s⁻¹/0.08 m

= 4250 Hz.

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

Ans: For Wave A:

Wavelength (λ A) = 2.5 cm=0.025 m

Frequency (ν A) = v/λ A=345/0.025=13800 Hz

For Wave B:

Wavelength (λ B) = 5.0 cm=0.05 m

Frequency (ν B ) = v/λ B=345/0.05=6900 Hz

15. 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?