Sound Test

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Sound Test - 42

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Question 1
1 / -0

The piston in a petrol engine goes up and down $$3000$$ times per minute. For this engine, calculate the period of the piston.

A
$$0.02\ s$$

B
$$0.04\ s$$

C
$$0.05\ s$$

D
$$0.08\ s$$

Solution

The number of complete oscillations per unit time is called frequency.
The time taken by the wave for one complete oscillation is called the time period,
Given,
Number of oscillations $$= 3000$$
Time for $$3000$$ oscillations, $$t=1\ min= 60\ s$$

Frequency, $$f=\dfrac{Number\ of\ oscillations}{time}$$

Frequency, $$f=\dfrac{3000}{60}$$

$$f=50\ Hz$$

Time period, $$T=\dfrac1f$$

$$T=\dfrac{1}{50}$$

$$T=0.02\ s$$
Question 2
1 / -0

A bicycle wheel spins 25 times in 5 seconds. Calculate frequency of the wheel.

A
15 Hz

B
10 Hz

C
40 Hz

D
5 Hz

Solution

Time Period (T) is the number of seconds per waveform, or the number of seconds per oscillation. It is clear that frequency and period are reciprocals.
In this case, A bicycle wheel spins 25 times in 5 seconds.
That is, the frequency is given as $$f=\frac { 25 }{ 5 } =5Hz$$.
Time period = $$\frac { 1 }{ f } =\frac { 1 }{ 5 } =0.20s$$
Hence, the frequency is 5 Hz and the time period is 0.20 seconds.
Question 3
1 / -0

The time period of a sound wave from a piano is $$1.18\times10^{-3} s$$. Find its frequency.

A
$$847.45 Hz$$

B
$$800 Hz$$

C
$$935. 55 Hz$$

D
$$1000 Hz$$

Solution

The frequency (f) of a wave is the number of full waveforms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.
Time Period (T) is the number of seconds taken to cover one wavelength or the number of seconds per oscillation. It is clear that frequency and period are reciprocals.
In this case, the time period is given as $$1.18\times { 10 }^{ -3 }s$$.
So, the frequency $$f=\frac { 1 }{ T } =\frac { 1 }{ 0.00118s } =847.45Hz.$$
Question 4
1 / -0

A tuning fork's tongs vibrate 250 times in 2.0s. Find the frequency of vibration

A
125 Hz

B
200 Hz

C
250 Hz

D
600 Hz

Solution

The frequency (f) of a wave is the number of full waveforms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.
Time Period (T) is the number of seconds per waveform or the number of seconds per oscillation. It is clear that frequency and period are reciprocals.
In this case, A tuning fork's tongs vibrate 250 times in 2.0 s.
That is, the frequency is given as $$f=\dfrac { 250 }{ 2 } =125 Hz$$.
Time period = $$\dfrac { 1 }{ f } =\dfrac { 1 }{ 125 } $$=$$8\times { 10 }^{ -3 }s$$.
Hence, the frequency is 125 Hz and the time period is $$8\times { 10 }^{ -3 }s$$ seconds.
Question 5
1 / -0

Which of the following pictures correctly show the way sound vibrations travel?

A

B

C

D

Solution

The answer is C.

When a body vibrates, the sound wave produced by it will travel in all directions. The third figure in the option depicts that the sound is travelling in all directions. Therefore, this figure correctly represents the travelling of sound.
Question 6
1 / -0

Calculate the period of a strobe light flashing $$25$$ times in $$5.0\ s$$.

A
$$5\ s$$

B
$$0.2\ s$$

C
$$3\ s$$

D
$$0.6\ s$$

Solution

The frequency $$f$$ of a wave is the number of cycles per second.
Time Period $$T$$ is defined as the time taken to complete one cycle.
The strobe light flashes $$25$$ times in $$5.0\ s$$.
$$f=\dfrac { 25 }{ 5 }$$
$$f =5\ Hz$$.

We know,
$$f=\dfrac{1}{T}$$

$$\implies T=\dfrac{1}{f}$$
Time period, $$T=\dfrac { 1 }{ 5 }$$
$$T =0.2\ s$$
Question 7
1 / -0

Calculate the frequency, in hertz, for 120 Oscillations in 2.0 s

A
60 Hz

B
100 Hz

C
80 Hz

D
120 Hz

Solution

The frequency (f) of a wave is the number of full waveforms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.
Time Period (T) is the number of seconds per waveform or the number of seconds per oscillation. It is clear that frequency and period are reciprocals.
In the first case, there are 120 Oscillations in 2.0 s.
That is, the frequency is given as $$f=\dfrac { 120 }{ 2 } =60Hz$$.
Question 8
1 / -0

Calculate the frequency for ten swings of a pendulum in $$6.7\ s$$.

A
$$1.492\ Hz$$

B
$$0.555\ Hz$$

C
$$2.372\ Hz$$

D
$$4.864\ Hz$$

Solution

The number of complete oscillations per unit time is called frequency.
Given,
Number of swings/oscillations $$= 10$$
Time for $$10$$ oscillations, $$t=6.7\ s$$

Frequency, $$f=\dfrac{Number\ of\ oscillations}{time}$$

Frequency, $$f=\dfrac{10}{6.7}$$

$$f=1.492\ Hz$$
Question 9
1 / -0

A scientist performed an experiment as shown in the picture above. What happened as air was pumped out of the jar and he rang the bell?

A
The sound became louder.

B
The sound became fainter first and then louder once all the air was pumped out.

C
The sound could not be heard anymore.

D
The sound was the same as before.

Solution

Sound waves need medium to propagate from one point to another such as a solid, liquid, or gas. The sound waves move through each of these mediums by vibrating the molecules in the matter. The molecules in solids are packed very tightly. Liquids are not packed as tightly as solids. And gases are very loosely packed. The spacing of the molecules enables sound to travel much faster through a solid than a gas. Sound travels about four times faster and farther in water than it does in air.
When air is pumped out from the jar by this process, vacuum will creat inside the jar which not allow to propagate sound wave.
Hence, when the bell rings, sound cannot be heard by anyone. (Option C)
Question 10
1 / -0

The depth of ocean at any place can be measured (estimated) with the help of

A
X rays

B
Ultrasonic waves

C
Radio waves

D
Ultraviolet rays

Solution

The depth of an ocean is estimated by a device , called SONAR , which sends ultrasonic waves towards the bottom of ocean , and after reflection, waves come back to the sonar. The time is recorded between the emitting and the receiving waves and having the speed of waves in water , we get the depth of ocean .