Sound Test

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Sound Test - 32

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Weekly Quiz Competition

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

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 3
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 4
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 5
1 / -0

Take a metallic tumbler and a tablespoon. Strike the tablespoon gently at the brim of the tumbler (as shown in the figure above). Now suspend a small thermocole ball touching the rim of the tumbler and touch it to the vibrating tumbler. See how far the ball is displaced. Now, again strike the tablespoon at the brim of the tumbler but strike it hard this time. Again touch the thermocole to the vibrating tumbler and see how far the ball is displaced. Choose the correct option.

A
Thermocole ball is displaced more when the tumbler is striked hard, as frequency if vibration is more in this case.

B
The thermocole ball is displaced less when the tumbler is striked hard, as frequency of vibration is less in this case.

C
The thermocole ball is displaced more then the tumbler is striked hard, as amplitude of vibration is more in this case.

D
The thermocole ball is displaced more when the tumbler is striked hard, as amplitude of vibration is less in this case.

Solution

In our observation, we see that the thermocole ball is displaced more when the tumbler is stricken hard, as the amplitude of vibration is more in this case.
The loudness of sound is proportional to the square of the amplitude of the vibration producing the sound. For example, if the amplitude becomes twice, the loudness increases by a factor of 4. The loudness is expressed in a unit called decibel (dB).
The loudness of sound depends on its amplitude. When the amplitude of vibration is large, the sound produced is loud. When the amplitude is small, the sound produced is feeble.
Hence, option C is correct.
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

Each of the four bowls below is filled with water up to different levels. When struck with a wooden rod, which bowl will produce sound of the lowest pitch?

A
a

B
b

C
c

D
d

Solution

Each of the bowls will have a different tone when hit with the wooden rod, the bowl with the most water will have the highest pitch while the bowl with the least water will have the lowest pitch. Small vibrations are made when you hit the bowl and this creates sound waves which travel through the water. Less water means faster vibrations, a lower frequency, that is lowest pitch.
Hence, when struck with a wooden rod, the bowl d will produce the sound of the lowest pitch.