Class 9 - Chapter 10: Sound Waves: Characteristics and Applications

Exercise Solution

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

Answer: (ii) Sound needs a medium to propagate

Mechanical waves require a material medium for transmission. Sound cannot travel through vacuum.

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

Answer: (iii) Number of compressions per second

Frequency is the number of vibrations or compressions produced per second.

3. If 20 compressions pass a point in 4 seconds, find the frequency.

Frequency (f) = Number of compressions / Time

f = 20 / 4

f = 5 Hz

Answer: (ii) 5 Hz

4. Reflected sound reaches the ear after 0.05 s. Will it produce an echo or reverberation?

For a distinct echo, the reflected sound must reach the ear after at least 0.1 s.

Given time = 0.05 s < 0.1 s

Therefore, a distinct echo will not be heard.

Answer: Reverberation

5. Compare wavelengths of the two waves in Fig. 10.30

Since both graphs have the same horizontal scale:

• Wave (a) has fewer cycles in the same distance → Greater wavelength.
• Wave (b) has more cycles in the same distance → Smaller wavelength.

Answer:

(i) Greater wavelength → Graph (a)

(ii) Smaller wavelength → Graph (b)

6. Identify A, B and C in Fig. 10.31

Frequency is inversely proportional to wavelength.

• Maximum frequency → Smallest wavelength
• Minimum frequency → Largest wavelength

From the figure:

• Green curve → Maximum frequency → A
• Blue curve → Minimum frequency → C
• Red curve → Intermediate frequency → B

7. Draw a graph representing a sound wave of amplitude 3 units and wavelength 4 cm.

The graph should:

• Have maximum displacement = +3 units
• Have minimum displacement = −3 units
• Complete one full cycle in 4 cm

Peak-to-peak distance = 4 cm.

8. Explosion in space shown with sound and light together

This is scientifically incorrect.

Light can travel through vacuum, but sound requires a medium.

Since outer space is nearly a vacuum, sound cannot travel there.

Error: The explosion should be visible but not audible.

9. Find the time period of the sound wave

Given:

Wavelength (λ) = 3.44 m

Speed (v) = 344 m s⁻¹

Frequency:

f = v / λ

f = 344 / 3.44

f = 100 Hz

Time period:

T = 1 / f

T = 1 / 100

T = 0.01 s

Answer: 0.01 s

10. Sonar problem

Speed of ultrasonic wave = 1525 m s⁻¹

Echo time = 5 s

Total distance travelled:

Distance = Speed × Time

= 1525 × 5

= 7625 m

Depth of ocean:

= 7625 / 2

= 3812.5 m

Answer: 3812.5 m

11. Parking sensor problem

Distance to obstacle = 1.2 m

Round-trip distance = 2 × 1.2

= 2.4 m

Speed of ultrasound = 345 m s⁻¹

Time:

t = Distance / Speed

t = 2.4 / 345

t = 0.00696 s

≈ 0.007 s

Answer: 0.007 s

12. Extra time taken by thunder

Distance = 1720 m

Time at 22°C:

t₁ = 1720 / 344 = 5 s

Time at 0°C:

t₂ = 1720 / 331

t₂ = 5.196 s

Extra time:

= 5.196 − 5

= 0.196 s

≈ 0.2 s

Answer: 0.2 s

13. Find wavelength and frequency (Fig. 10.32)

The marked 8 cm distance contains two complete wavelengths.

Therefore:

λ = 8 / 2

λ = 4 cm

= 0.04 m

Given speed:

v = 340 m s⁻¹

Frequency:

f = v / λ

f = 340 / 0.04

f = 8500 Hz

Answer:

Wavelength = 4 cm

Frequency = 8500 Hz

14. Find wavelength and frequency of waves A and B (Fig. 10.33)

From the graph:

• Wave A completes about 2.5 cycles in 7.5 cm.
• Wave B completes about 1.5 cycles in 7.5 cm.

Wave A:

λ_A = 7.5 / 2.5 = 3 cm = 0.03 m

f_A = 345 / 0.03

f_A = 11500 Hz

Wave B:

λ_B = 7.5 / 1.5 = 5 cm = 0.05 m

f_B = 345 / 0.05

f_B = 6900 Hz

Answer:

Wave A: λ = 3 cm, f = 11500 Hz

Wave B: λ = 5 cm, f = 6900 Hz

15. Ratio of speeds of sound in air and water

Let:

Distance to cliff = d

Speed in air = v_a

Speed in water = v_w

Round-trip time:

t_a = 2d / v_a

t_w = 2d / v_w

Given:

t_a = 4.5 t_w

Therefore:

2d / v_a = 4.5(2d / v_w)

v_w / v_a = 4.5

Hence:

v_a : v_w = 1 : 4.5

Answer: 1 : 4.5