# Question-1

**How does the sound produced by a vibrating object in a medium reach your ear?**

**Solution:**

The vibrating object produces a series of compressions and rarefactions, one after the other in the medium. These pulses travel one behind the other as the sound waves move forward. When this sound wave reaches our ear, it forces the tympanic membrane to vibrate and thus causes the sensation of hearing.

# Question-2

**Explain how sound is produced by your school bell.**

**Solution:**

When a hammer hits the school bell, it begins to vibrate and produce sound. If we touch the bell gently just after hitting it, we will feel these vibrations. We can say that sound is produced by a vibrating body.

# Question-3

**Why are sound waves called mechanical waves?**

**Solution:**

Sound needs a material medium like air, water, steel, etc. for its propagation. It is characterized by the motion of the particles of the medium. So, sound waves are called mechanical waves.

# Question-4

**Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?**

**Solution:**

No, the moon has no atmosphere or material medium through which sound can

Â travel.

# Question-5

**What is a wavelength, a frequency, a time period and the amplitude of a soundÂ**

wave?

wave?

**Solution:**

Wavelength (l ): It is defined as the distance between two consecutive compressions or rarefactions of a sound wave.

Frequency (n ): The number of complete oscillations or cycles per unit time is called frequency.

Time period (T): The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period of the sound wave.

Amplitude (A): The magnitude of the maximum disturbance in the medium on either side of the mean value is called the amplitude of the wave.

# Question-6

**How are the wavelength and frequency of a sound wave related to its speed?**

**Solution:**

Speed of the wave = Frequency Ã— Wavelength Or Ï… = Î½ Î»

# Question-7

**Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 ms**

^{-1}in a given medium.**Solution:**

Wavelength = = 2 m

Â

# Question-8

**A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of sound. What is the time interval between successive compressions from the source?**

**Solution:**

Time interval between two successive compressions,

T = = 0.002 s.

Â

# Question-9

**Which wave property determines (a) loudness, b) pitch?**

**Solution:**

(a) Amplitude of the wave determines the loudness.

(b) Frequency of the wave determines the pitch.

# Question-10

**Guess which sound has a higher pitch guitar or car horn?**

**Solution:**

The sound of a guitar is shriller than that of a car horn. So, the sound of a guitar has a higher pitch.

# Question-11

**Distinguish between loudness and intensity of sound.**

**Solution:**

# Question-12

**In which of the three media: air, water or iron, does sound travel the very fast at a particular temperature?**

**Solution:**

Sound travels fastest through iron with a speed of 5950 ms

^{-1}.

# Question-13

**An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms**

^{-1}?**Solution:**

Speed of sound, Ï… = 342 ms

^{-1}

Time taken for hearing the echo, t = 3 s

Distance traveled by the sound = Ï… t = 342 Ã— 3 = 1026 m

In 3s, sound has to travel twice the distance between the reflecting surface and the source.

âˆ´ Distance of the reflecting surface from the source = = 513 m.

# Question-14

**Why are the ceilings of the concert halls curved?**

**Solution:**

The ceilings of the concert halls are curved so that sound after reflecting from the ceiling reaches all corners of the hall.

Â

# Question-15

**What is the audible range of the an average human ear?**

**Solution:**

For an average human ear, the audible range of frequency extends form 20 Hz to 20,000 Hz.

# Question-16

**What is the range of frequencies associated with (a) Infrasound? (b) Ultrasound?**

**Solution:**

(a) Sound of frequency which is less than 20 Hz is called infrasound.

(b) Sound of frequency which is higher than 20 kHz is called ultrasound.

# Question-17

**A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?**

**Solution:**

Time between transmission and detection of sonar pulse, t = 1.02 s

Speed of sound in salt water, Ï… = 1531 ms

^{-1}

Distance of the cliff = d (say)

Then the distance traveled by sound = 2d

But 2d = Speed of sound Ã— time = Ï… t

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1531 Ã— 1.02 m

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â d = = 780.81 m.

# Question-18

**What is the time period of a pendulum of length 9.8 metres ?**

**Solution:**

Here,

Length of pendulam, L = 9.8m

Time period, T= ? (to be calculated)

The time period is 6.28 s.

# Question-19

**The length of a simple pendulum is 84 cm. What will be its periodic time if it were taken to the Moon,where the acceleration due to gravity is one-sixth that on the Earth? (Take the value of g on Earth = 9.8 m/s**

^{2}).**Solution:**

Value of g on the Earth = 9.8 m/s

^{2}

Value of g on Moon = m/s

^{2 }

Length of the pendulum = 84 cm = 0.84 m

Time period T = 2Ï€

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 Ã— 3.14 Ã— = 4.5s.

# Question-20

**From the graph find the distances between the following pairs of points in terms of wavelength**

**l**

**:**

(a) E and F (b) F and G

(c) E and G (d) F and H

(e) G and H (f) E and H

(a) E and F (b) F and G

(c) E and G (d) F and H

(e) G and H (f) E and H

**Solution:**

(a) EF = Î» /4

(b) FG = Î» /4

(c) EG = Î» /2

(d) FH = Î» /2

(e) GH = Î» /4

(f) EH = 3Î» /4.

# Question-21

**A bat can hear sounds at frequencies upto 120 kHz. Determine the wavelength of sound in air at that frequency. Take the speed of sound in air as 344 m/s.**

**Solution:**

Frequency, Î½ = 120 kHz = 120000 Hz

Speed of sound in air, v = 344 m/s

We know that v = Î½ Î»

âˆ´ Wavelength, Î» = = 344 Ã— = 0.29 cm/s.

Â

# Question-22

**We know that at the surface of the Moon there is no atmosphere. Suppose you and your friend land on the Moon, would you and your friend be able to talk to each other? Why?**

**Solution:**

No, because the moon has no atmosphere or medium through which sound can travel.

# Question-23

**The time period of a simple pendulum, for small swings, depends on**

(a) the mass of the body,

(b) its length,

(c) the size of the bob,

(d) its amplitude.

(a) the mass of the body,

(b) its length,

(c) the size of the bob,

(d) its amplitude.

**Solution:**

(b) its length.

# Question-24

**The speed of the bob of an oscillating pendulum is maximum**

(a) at each of the extreme positions,

(b) between the mean position and the right extreme position,

(c) between the mean position and the left extreme position,

(d) at the mean position.

(a) at each of the extreme positions,

(b) between the mean position and the right extreme position,

(c) between the mean position and the left extreme position,

(d) at the mean position.

**Solution:**

(d) at the mean position.

# Question-25

**If the length of a simple pendulum is doubled, its time period**

(a) is halved,

(b) is doubled,

(c) becomes âˆš2 times,

(d) reduces by âˆš2

(a) is halved,

(b) is doubled,

(c) becomes âˆš2 times,

(d) reduces by âˆš2

**.**

**Solution:**

(c) becomes âˆš2 times.

# Question-26

**A human heart, on an average, is found to beat 75 times a minute. Calculate its frequency.**

**Solution:**

Frequency = = 1.25 /s.

# Question-27

**Find the length of a pendulum which will have the same time period on the Moon as that of a pendulum of length 96 cm on the Earth. The value of acceleration due to gravity on the moon is one-sixth of that of the earth.**

**Solution:**

For a simple pendulum,

T = 2Ï€

and

Time period T is kept constant, therefore

T = Â Â Â Â Â Â Â Â Â Â Â Â Â ----------(1)

T = Â Â Â Â Â Â Â Â Â Â Â ----------(2)

Dividing equation (1) by (2), we get

1 =

1 =

âˆ´ = 16 cm.

# Question-28

**What is the time period of a pendulum of length 9.8 metres?**

**Solution:**

Time period T = 2Ï€

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â T = 2Ï€

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2Ï€ s.

# Question-29

**Which of the following is carried by the waves from one place to another?**

(a) massÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) velocity

(c) wavelengthÂ Â Â Â Â Â Â (d) energy.

(a) massÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (b) velocity

(c) wavelengthÂ Â Â Â Â Â Â (d) energy.

**Solution:**

(d) energy.

# Question-30

**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 the speed of sound in air as 340 m/s.**

**Solution:**

Velocity v = Î½ Î»

Given v = 340 m/s

Î½ = 20 Hz

âˆ´Î» = = 17 m

If Î½ = 20 kHz

then Î» = = 0.007 m.

# Question-31

**Two children are at opposite ends of an iron pipe. One strikes an end of the pipe with a stone. Find the ratio of time taken by the sound waves in air and in iron, each, to reach the other child. Velocity is sound wave in air = 344 m/s and in**

**iron = 5130 m/s.**

**Solution:**

Velocity of sound in air = 344 m/s

Velocity of sound in iron = 5130 m/s

âˆ´ Ratio =

**Â Â Â Â Â Â Â Â Â Â Â Â Â**= 15.

# Question-32

**A boat at anchor is rocked by waves whose consecutive crests are 100 m apart. The wave velocity of the moving crests is 20 m/s. What is the frequency of rocking of the boat?**

**Solution:**

Wavelength Î» = 100 m

Wave velocity, v = 20 m/s

âˆ´ Î½ = = = 0.2 s

^{-1}.

Â

# Question-33

**A longitudinal wave is produced on a toy slinky. The wave travels at a speed of 30 cm/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions of slinky?**

**Solution:**

Velocity v = 30 cm/s

Frequency, Î½ = 20 Hz

âˆ´ Wavelength Î» =

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = = 1.5 m.

Â

# Question-34

**A source of wave produces 40 crests and 40 troughs in 0.4 s. Find the frequency of the wave.**

**Solution:**

One complete wave cycle consists of one crest and one trough.

The given source produces 40 crests, hence it produces 40 wave cycles in the given interval 0.4 s.

âˆ´ Number of wave cycles produced in 1s = = 100

We know that the number of wave cycles produced in one second is equal to the frequency.

Therefore the frequency of the wave is 100 Hz.

# Question-35

**Using SONAR, sound pulses are emitted from the surface of the ocean and these pulses after being reflected from the bottom are detected. If the time interval from the emission to the detection of the sound pulse in 2 second, find the depth of the water.**

**Solution:**

Depth of the water, d =

Time interval t = 2 s

Velocity of the sound waves in water, v = 1500 m/s

âˆ´ d = = 1500 m.

Â

# Question-36

**A child hears an echo from a cliff 4 seconds after the sound from a powerful cracker is produced.How far away is the cliff from the child? Velocity of sound in air is 350 m/s.**

**Solution:**

Distance travelled by the sound = 2d

We know that v =

âˆ´ d = = 700m.

Â

# Question-37

**A stone is dropped into a well, 44.1 metres deep. The sound of the splash is heard 3.13 seconds after the stone is dropped. Find the velocity of sound in air.**

**Solution:**

Depth of the well, d = 44.1 m

Time taken to hear the splash, t = 3.13 s

Let t

_{1}be the time taken by the stone to hit the water surface in the well and t

_{2}be the time taken by the splash of sound to reach the top of the well.

âˆ´ t

_{1}+ t

_{2}= 3.13 s ---------------- (1)

When the stone moves down,

Velocity u = 0

Acceleration a = 9.8 m/s

^{2}

Distance travelled = 44.1 m

Using the formula

s = ut + at

^{2}

44.1 = 0 + Ã— 9.8 t

^{2}

= 4.9 t

^{2}

= 9

âˆ´ t

_{1}= = 3s

From the equation (1), we get

3 + t

_{2}= 3.13

âˆ´ t

_{2}= 3.13 â€“ 3 = 0.13 s

If c is the velocity of sound in air, then

c = = 340 m/s.