Q1. What is sound and how is it produced?

Answer: Sound is a form of energy created by vibrating objects. As an object vibrates, it produces longitudinal waves—compressions and rarefactions—in a material medium, which propagate to our ears as sound.

Q2. Describe with the help of a diagram how compressions and rarefactions are produced in air near a source of sound.

Answer: When a tuning fork prong moves outward, it compresses adjacent air molecules (compression). As it moves inward, it creates a low-pressure region (rarefaction). These alternate along the wave path.

Q3. Cite an experiment to show that sound needs a material medium for its propagation.

Answer: Place a ringing bell under a vacuum bell jar. As air is pumped out, the bell’s sound fades and stops, demonstrating that sound cannot travel in a vacuum.

Q4. Why is sound wave called a longitudinal wave?

Answer: In sound waves, particles of the medium oscillate parallel to the direction of wave propagation, characteristic of longitudinal waves.

Q5. Which characteristic of sound helps you identify your friend by his voice in a dark room?

Answer: Quality (timbre), which depends on the waveform and harmonic content of the voice, allows recognition of different voices.

Q6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Answer: Light travels much faster (3×10⁸ m/s) than sound (≈340 m/s). The flash reaches us almost instantly, while thunder’s sound wave arrives later.

Q7. A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take speed of sound = 344 m/s.
Answer:

Given:
Lowest frequency,

$f_1 = 20 \, \text{Hz}$

Highest frequency,

$f_2 = 20,000 \, \text{Hz}$

Speed of sound,

$v = 344 \, \text{m/s}$

Wavelengths,

$\lambda_1, \lambda_2 = ?$

Formula:

$$\lambda = \frac{v}{f}$$

Calculating wavelength for lowest frequency:

$$\lambda_1 = \frac{344}{20} = 17.2 \, \text{m}$$

Calculating wavelength for highest frequency:

$$\lambda_2 = \frac{344}{20,000} = 0.0172 \, \text{m}$$

Typical wavelength range of sound waves in air = 0.0172 m to 17.2 m.

Q8. Two children are at opposite ends of an aluminium rod. One strikes the end with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Answer:

Given:
Speed of sound in air,

$v_{\text{air}} = 340 \, \text{m/s}$

(approximate)
Speed of sound in aluminium,

$v_{\text{Al}} = 5000 \, \text{m/s}$

(approximate)
Length of rod (distance between children),

$d$

(same for both media)
Time taken in air,

$t_{\text{air}} = \frac{d}{v_{\text{air}}}$

Time taken in aluminium,

$t_{\text{Al}} = \frac{d}{v_{\text{Al}}}$

Ratio of times:

$$\frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{\frac{d}{v_{\text{air}}}}{\frac{d}{v_{\text{Al}}}} = \frac{v_{\text{Al}}}{v_{\text{air}}}$$

Substitution:

$$\frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{5000}{340} \approx 14.7$$

The sound takes approximately 14.7 times longer to travel through air than through aluminium.

Q9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Answer:

Given Frequency = 100 Hz.

Number of vibration in one minute = 100 vibrations/s × 60 s = 6000 vibrations.

Q10. Does sound follow the same laws of reflection as light does? Explain.

Answer: Yes. The angle of incidence equals the angle of reflection for sound waves at a rigid boundary, following the same geometric rules as light.

Q11. When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source remain the same. Do you hear echo sound on a hotter day?

Answer: Yes, but with a shorter delay. Speed of sound increases with temperature, so the echo returns slightly sooner.

Q12. Give two practical applications of reflection of sound waves.

Answer: Megaphones to direct sound towards an audience; sonar to locate underwater objects.

Q13. A stone is dropped from a tower 500 m high into a pond. When is the splash heard at the top? Given

$g = 10 \, \text{m/s}^2$

and speed of sound

$v = 340 \, \text{m/s}$

.
Answer:

Given:
Height of tower,

$h = 500 \, \text{m}$

Acceleration due to gravity,

$g = 10 \, \text{m/s}^2$

Speed of sound,

$v = 340 \, \text{m/s}$

Time taken for stone to fall,

$t_1 = ?$

Time taken for sound to travel up,

$t_2 = ?$

Total time for splash to be heard,

$T = t_1 + t_2$

Calculating time taken by stone to reach the pond using:

$$h = \frac{1}{2} g t_1^2$$

$$500 = \frac{1}{2} \times 10 \times t_1^2$$

$$500 = 5 t_1^2$$

$$t_1^2 = \frac{500}{5} = 100$$

$$t_1 = \sqrt{100} = 10 \, \text{seconds}$$

Calculating time taken by sound to travel up:

$$t_2 = \frac{h}{v} = \frac{500}{340} \approx 1.47 \, \text{seconds}$$

Total time:

$$T = t_1 + t_2 = 10 + 1.47 = 11.47 \, \text{seconds}$$

The splash is heard at the top after approximately 11.47 seconds.

Q14. A sound wave travels at 339 m/s. If its wavelength is 1.5 cm, what is the frequency? Is it audible?

Answer: f = v/λ = 339 m/s ÷ 0.015 m ≈22600 Hz. Slightly above 20 kHz, just beyond typical hearing range.

Q15. What is reverberation? How can it be reduced?

Answer: Reverberation is persistence of sound due to multiple reflections in an enclosure. It is reduced by adding sound-absorbing materials (curtains, carpets, foam panels).

Q16. What is loudness of sound? What factors does it depend on?

Answer: Loudness is the perceived intensity of sound, depending on amplitude, frequency, duration, and listener sensitivity.

Q17. Explain how bats use ultrasound to catch prey.

Answer: Bats emit ultrasonic pulses and detect returning echoes from insects. Echo timing and intensity provide distance and location of prey, enabling capture in darkness.

Q18. How is ultrasound used for cleaning?

Answer: Objects immersed in a cleaning solution are subjected to ultrasonic vibrations, creating cavitation bubbles that collapse and dislodge dirt and contaminants from surfaces, including crevices.

Q19. Explain the working and application of sonar.
Answer:

Working of Sonar:
Sonar (Sound Navigation and Ranging) works by emitting ultrasonic sound waves from a transmitter in a submarine or ship. These sound waves travel through water, hit an object (like the ocean floor or another underwater object), and get reflected back as echoes. The sonar receiver detects these echoes. By measuring the time interval between sending the sound wave and receiving its echo, the distance to the object is calculated using the formula:

$$\text{Distance} = \frac{\text{Speed of sound in water} \times \text{Time taken}}{2}$$

Dividing by 2 accounts for the sound wave’s journey to the object and back.

Applications of Sonar:

1. Navigation and Detection: Helps submarines and ships detect underwater objects, obstacles, or other vessels.
2. Depth Measurement: Used to measure the depth of oceans, lakes, and rivers.
3. Fishing: Helps fishermen locate schools of fish by detecting underwater fish movements.
4. Scientific Research: Used for mapping the ocean floor and studying marine life habitats.

Sonar is essential for safe underwater navigation and exploration.

Q20. A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the object is 3625 m away.
Answer:

Given:
Time for echo to return,

$t = 5$

s
Distance to object,

$d = 3625$

m
Speed of sound in water,

$v = ?$

Formula:

$$\text{Total distance travelled by sound} = 2d = v \times t$$

Rearranged for speed:

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

Substitution:

$$v = \frac{2 \times 3625}{5} = \frac{7250}{5} = 1450 \, \text{m/s}$$

Speed of sound in water = 1450 m/s.