Class 9th Science Chapter-11 Exercises

Exercises – Chapter – 11: Sound

Questions with Answers

---

Question 1

What is sound and how is it produced?

Answer:
Sound is a form of energy that produces the sensation of hearing in our ears.
Sound is produced when an object vibrates. These vibrations set the particles of the surrounding medium into motion, producing sound.

---

Question 2

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

Answer:
When a vibrating object moves forward, it compresses the air in front of it, creating a region of high pressure called compression.
When it moves backward, it creates a region of low pressure called rarefaction.
As the object vibrates continuously, a series of compressions and rarefactions is formed, which travels through air as a sound wave.

---

Question 3

Why is sound wave called a longitudinal wave?

Answer:
Sound wave is called a longitudinal wave because the particles of the medium vibrate back and forth parallel to the direction of propagation of the wave.

---

Question 4

Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Answer:
The quality (or timbre) of sound helps us to identify a person by his or her voice.

---

Question 5

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

Answer:
Light travels much faster than sound.
Therefore, the flash of lightning reaches our eyes almost instantly, while the sound of thunder takes more time to reach our ears.

---

Question 6

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 in air $v=344m{s}^{-1}$)

Answer:

Using the relation:

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

For $\nu =20Hz$:

$$\lambda =\frac{344}{20}=17.2m$$

For $\nu =20000Hz$:

$$\lambda =\frac{344}{20000}=0.0172m$$

---

Question 7

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

Answer:

Speed of sound in air = 344 m s⁻¹
Speed of sound in aluminium = 6420 m s⁻¹

Time taken $t\propto \frac{1}{v}$

$$\text{Ratio of times}=\frac{v_{\text{aluminium}}}{v_{\text{air}}}=\frac{6420}{344}\approx 18.7$$

Ratio = 18.7 : 1

---

Question 8

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

Answer:
Frequency = number of vibrations per second

In 1 second = 100 vibrations
In 60 seconds:

$$100\times 60=6000$$

Number of vibrations = 6000

---

Question 9

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

Answer:
Yes, sound follows the same laws of reflection as light.

Explanation:

• The angle of incidence is equal to the angle of reflection.

• The incident sound, reflected sound and the normal lie in the same plane.

---

Question 10

When a sound is reflected from a distant object, an echo is produced. If the distance between the reflecting surface and the source remains the same, do you hear echo sound on a hotter day?

Answer:
Yes, the echo is heard earlier on a hotter day.

Explanation:
On a hotter day, the speed of sound increases, so the reflected sound returns faster.

---

Question 11

Give two practical applications of reflection of sound waves.

Answer:

1. Megaphones and loudhailers are designed using reflection of sound to direct sound forward.

2. Stethoscope uses multiple reflections of sound to transmit heartbeats clearly to the doctor’s ears.

---

Question 12

A stone is dropped from the top of a tower 500 m high into a pond at the base. When is the splash heard at the top?
(Given $g=10m{s}^{-2}$, speed of sound $=340m{s}^{-1}$

Answer:

Time taken by stone to fall:

$$s=\frac{1}{2}g{t}^2\Rightarrow 500=5t^2\Rightarrow t=10s$$

Time taken by sound to travel up:

$$t=\frac{500}{340}\approx 1.47s$$

Total time:

$$10+1.47=11.47s$$

The splash is heard after approximately 11.5 s.

---

Question 13

A sound wave travels at a speed of 339 m s⁻¹. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Answer:

Given:
Speed, $v=339m{s}^{-1}$
Wavelength, $\lambda =1.5cm=0.015m$

$$\nu =\frac{v}{\lambda }=\frac{339}{0.015}=22600Hz$$

The audible range for humans is 20 Hz to 20 kHz.

Therefore, the sound will not be audible, as its frequency is greater than 20 kHz (ultrasound).

---

Question 14

What is reverberation? How can it be reduced?

Answer:
Reverberation is the persistence of sound in a large hall due to multiple reflections from walls, ceiling and other surfaces, even after the source has stopped producing sound.

It can be reduced by:

• Covering walls and ceilings with sound-absorbing materials such as fibreboard or rough plaster

• Using curtains, carpets and cushioned seats

---

Question 15

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

Answer:
Loudness is the physiological response of the human ear to the sound.

It depends on:

• The amplitude of the sound wave

• The sensitivity of the human ear

A sound with larger amplitude is heard as louder.

---

Question 16

How is ultrasound used for cleaning?

Answer:
In ultrasonic cleaning, objects are placed in a liquid and ultrasonic waves are passed through it.
These high-frequency waves produce vibrations that dislodge dust, grease and dirt from even hard-to-reach places.
Thus, the objects get thoroughly cleaned.

---

Question 17

Explain how defects in a metal block can be detected using ultrasound.

Answer:
Ultrasonic waves are passed through the metal block.
If there is a defect or crack inside, the waves get reflected back from that place.
By detecting the reflected waves, the presence and location of defects in the metal block can be identified.