Class 9 Science NCERT Solution chapter 10 Sound Waves — Characteristics and Applications

Revise, Reflect, Refine — Complete Solutions

---

🟦 QUESTION 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.

🟩 ANSWER

Correct option: (ii) Sound needs a medium to propagate.

A mechanical wave requires a material medium for its propagation. Sound travels through vibrations of particles in solids, liquids or gases.

Sound cannot travel through a vacuum because there are no particles to transfer the disturbance.

---

🟦 QUESTION 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

🟩 ANSWER

Correct option: (iii) Number of compressions per second

Frequency is the number of complete oscillations or compressions passing a point per second.$$\nu=\frac{\text{Number of oscillations}}{\text{Time}}$$ν=TimeNumber of oscillations​

Therefore, when frequency increases, the number of compressions passing per second also increases.

For sound travelling through the same medium:

• Speed remains nearly constant.
• Wavelength decreases because $v=\lambda\nu$v=λν.
• Time period decreases because $T=\frac{1}{\nu}$T=ν1​.

---

🟦 QUESTION 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

🟩 ANSWER

Correct option: (ii) 5 Hz

Frequency is:$$\nu=\frac{\text{Number of compressions}}{\text{Time}}$$ν=TimeNumber of compressions​$$\nu=\frac{20}{4}$$ν=420​$$\boxed{\nu=5\text{ Hz}}$$ν=5 Hz​

Therefore, five compressions pass the point every second.

---

🟦 QUESTION 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.

🟩 ANSWER

It will produce reverberation, not a distinct echo.

For a reflected sound to be heard separately as an echo, the time interval between the original and reflected sound should be at least about:$$0.1\text{ s}$$0.1 s

Here, the reflected sound reaches the ear after only:$$0.05\text{ s}$$0.05 s

Therefore, the reflected sound merges with the original sound and causes persistence or prolongation of sound, known as reverberation.

---

🟦 QUESTION 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 sound wave has:

(i) Greater wavelength?
(ii) Smaller amplitude?

🟩 ANSWER

According to Fig. 10.30 on page 205:

(i) Greater wavelength

Wave (a) has the greater wavelength.

It completes fewer oscillations over the same distance. Therefore, the distance between two consecutive crests or troughs is greater.

(ii) Smaller amplitude

Wave (a) has the smaller amplitude.

Its crests and troughs show a smaller maximum displacement from the mean-density line than wave (b).

Final Answer:

• Greater wavelength — Wave (a)
• Smaller amplitude — Wave (a)

---

🟦 QUESTION 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 that of C is minimum, identify the corresponding curves and mark A, B and C on them.

🟩 ANSWER

Frequency depends on the number of complete oscillations occurring over a given distance.

From Fig. 10.31 on page 205:

• The green curve completes the maximum number of oscillations. Therefore, it represents A.
• The red curve has an intermediate number of oscillations. Therefore, it represents B.
• The blue curve completes the minimum number of oscillations. Therefore, it represents C.

Source	Curve	Frequency
A	Green curve	Maximum
B	Red curve	Intermediate
C	Blue curve	Minimum

The curve representing A also has the shortest wavelength, while the curve representing C has the longest wavelength.

---

🟦 QUESTION 7

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

🟩 ANSWER

Take:

• Distance in centimetres on the x-axis.
• Change in density on the y-axis.
• Mark the mean-density position as zero.
• Maximum density change or amplitude = $+3$+3 units.
• Minimum density change = $-3$−3 units.
• Wavelength = $4\text{ cm}$4 cm.

Since one complete wave occupies 4 cm, the quarter wavelength is:$$\frac{\lambda}{4}=\frac{4}{4}=1\text{ cm}$$4λ​=44​=1 cm

Plot the following points:

Distance (cm)	Density change (units)
0	0
1	+3
2	0
3	−3
4	0
5	+3
6	0
7	−3
8	0

Join these points using a smooth wave-like curve.

The distance between two successive crests, such as at 1 cm and 5 cm, is:$$5-1=4\text{ cm}$$5−1=4 cm

Therefore, the wavelength is correctly represented as 4 cm.

---

🟦 QUESTION 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?

🟩 ANSWER

There are two scientific errors in this depiction:

1. Sound cannot travel through space.
   Space is nearly a vacuum and contains no material medium through which sound waves can propagate. Therefore, the sound of the explosion cannot directly reach an observer.
2. Light and sound would not reach at the same time even if a medium were present.
   Light travels much faster than sound. Therefore, the flash would be seen before the sound was heard.

In real space, an observer would see the flash but would not hear the explosion unless its vibrations were transmitted through a spacecraft or converted into electrical signals.

---

🟦 QUESTION 9

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.

🟩 ANSWER

Given:$$\lambda=3.44\text{ m}$$λ=3.44 m$$v=344\text{ m s}^{-1}$$v=344 m s−1

Using:$$v=\lambda\nu$$v=λν$$\nu=\frac{v}{\lambda}$$ν=λv​$$\nu=\frac{344}{3.44}$$ν=3.44344​$$\nu=100\text{ Hz}$$ν=100 Hz

Time period:$$T=\frac{1}{\nu}$$T=ν1​$$T=\frac{1}{100}$$T=1001​$$\boxed{T=0.01\text{ s}}$$T=0.01 s​

Final Answer: Time period = 0.01 s

---

🟦 QUESTION 10

A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If the 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 located?

🟩 ANSWER

The given time is the total time taken by the sound to travel to the wreckage and return to the ship.

Given:$$v=1525\text{ m s}^{-1}$$v=1525 m s−1$$t=5\text{ s}$$t=5 s

Distance to the wreckage:$$d=\frac{v\times t}{2}$$d=2v×t​$$d=\frac{1525\times5}{2}$$d=21525×5​$$d=\frac{7625}{2}$$d=27625​$$d=3812.5\text{ m}$$d=3812.5 m$$\boxed{d\approx3.81\text{ km}}$$d≈3.81 km​

Final Answer: The wreckage is approximately 3812.5 m or 3.81 km below the ship.

---

🟦 QUESTION 11

A vehicle is fitted with an ultrasonic distance sensor as part of a parking-assistance system. The sensor emits an ultrasonic wave of about 40 kHz, which is reflected by an obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by the ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasound in air to be 345 m s−1345\text{ m s}^{-1}345 m s−1.

🟩 ANSWER

Distance between the sensor and obstacle:$$d=1.2\text{ m}$$d=1.2 m

The wave travels to the obstacle and returns.

Total distance travelled:$$2d=2\times1.2=2.4\text{ m}$$2d=2×1.2=2.4 m

Using:$$t=\frac{\text{Distance}}{\text{Speed}}$$t=SpeedDistance​$$t=\frac{2.4}{345}$$t=3452.4​$$t\approx0.00696\text{ s}$$t≈0.00696 s$$\boxed{t\approx0.007\text{ s}}$$t≈0.007 s​

In milliseconds:$$0.007\text{ s}=7\text{ ms}$$0.007 s=7 ms

Final Answer: The ultrasonic wave takes approximately 0.007 s or 7 milliseconds.

---

🟦 QUESTION 12

The speed of sound in air is about 331 m s−1331\text{ m s}^{-1}331 m s−1 at 0∘C0^\circ\text{C}0∘C and nearly 344 m s−1344\text{ m s}^{-1}344 m s−1 at 22∘C22^\circ\text{C}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∘C22^\circ\text{C}22∘C to 0∘C0^\circ\text{C}0∘C? Assume all other conditions remain unchanged.

🟩 ANSWER

Time taken at 22∘C22^\circ\text{C}22∘C

$$t_{22}=\frac{1720}{344}$$t22​=3441720​$$t_{22}=5\text{ s}$$t22​=5 s

Time taken at 0∘C0^\circ\text{C}0∘C

$$t_0=\frac{1720}{331}$$t0​=3311720​$$t_0\approx5.20\text{ s}$$t0​≈5.20 s

Extra time:$$\Delta t=t_0-t_{22}$$Δt=t0​−t22​$$\Delta t=5.20-5.00$$Δt=5.20−5.00$$\boxed{\Delta t\approx0.20\text{ s}}$$Δt≈0.20 s​

Final Answer: The thunder will take approximately 0.2 s longer at 0∘C0^\circ\text{C}0∘C.

---

🟦 QUESTION 13

The variation of density of a 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.

🟩 ANSWER

In Fig. 10.32 on page 206, the marked distance of 8 cm covers two complete wavelengths, from one compression to the third compression.

Therefore:$$2\lambda=8\text{ cm}$$2λ=8 cm$$\lambda=\frac{8}{2}$$λ=28​$$\lambda=4\text{ cm}$$λ=4 cm

Convert into metres:$$\lambda=0.04\text{ m}$$λ=0.04 m

Speed of sound:$$v=340\text{ m s}^{-1}$$v=340 m s−1

Using:$$v=\lambda\nu$$v=λν$$\nu=\frac{v}{\lambda}$$ν=λv​$$\nu=\frac{340}{0.04}$$ν=0.04340​$$\boxed{\nu=8500\text{ Hz}}$$ν=8500 Hz​

Final Answer:

• Wavelength = 4 cm or 0.04 m
• Frequency = 8500 Hz or 8.5 kHz

---

🟦 QUESTION 14

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 wave? Also calculate their frequencies.

🟩 ANSWER

According to Fig. 10.33 on page 206, the distance shown from 0 to the end of the graph is 7.5 cm.

Wave A

Wave A completes three full oscillations in 7.5 cm.$$\lambda_A=\frac{7.5}{3}$$λA​=37.5​$$\lambda_A=2.5\text{ cm}$$λA​=2.5 cm$$\lambda_A=0.025\text{ m}$$λA​=0.025 m

Frequency:$$\nu_A=\frac{v}{\lambda_A}$$νA​=λA​v​$$\nu_A=\frac{345}{0.025}$$νA​=0.025345​$$\boxed{\nu_A=13{,}800\text{ Hz}}$$νA​=13,800 Hz​

Wave B

Wave B completes 1.5 oscillations in 7.5 cm.$$\lambda_B=\frac{7.5}{1.5}$$λB​=1.57.5​$$\lambda_B=5\text{ cm}$$λB​=5 cm$$\lambda_B=0.05\text{ m}$$λB​=0.05 m

Frequency:$$\nu_B=\frac{345}{0.05}$$νB​=0.05345​$$\boxed{\nu_B=6900\text{ Hz}}$$νB​=6900 Hz​

Final Answer:

Wave	Wavelength	Frequency
A	2.5 cm or 0.025 m	13,800 Hz or 13.8 kHz
B	5 cm or 0.05 m	6,900 Hz or 6.9 kHz

Wave A has a higher frequency because it has a shorter wavelength.

---

🟦 QUESTION 15

Two identical sound sources are placed at A and B—one in air and one submerged in water, as shown in 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 sound to return to A is 4.5 times that taken to return to B, what is the ratio between the speeds of sound in air and water?

🟩 ANSWER

Both sound waves travel the same total distance to the cliff and back.

Let:

• Speed of sound in air = $v_a$va​
• Speed of sound in water = $v_w$vw​
• Time taken in air = $t_a$ta​
• Time taken in water = $t_w$tw​

Given:$$t_a=4.5t_w$$ta​=4.5tw​

For the same distance:$$v=\frac{d}{t}$$v=td​

Therefore:$$\frac{v_a}{v_w}=\frac{t_w}{t_a}$$vw​va​​=ta​tw​​$$\frac{v_a}{v_w}=\frac{t_w}{4.5t_w}$$vw​va​​=4.5tw​tw​​$$\frac{v_a}{v_w}=\frac{1}{4.5}$$vw​va​​=4.51​

Multiplying by 2:$$\frac{v_a}{v_w}=\frac{2}{9}$$vw​va​​=92​$$\boxed{v_{\text{air}}:v_{\text{water}}=2:9}$$vair​:vwater​=2:9​

Final Answer: The ratio of the speed of sound in air to that in water is 2:92:92:9.