Tag: time period, frequency and amplitude of sound

Questions Related to time period, frequency and amplitude of sound

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

  1. $847.45 Hz$

  2. $800 Hz$

  3. $935. 55 Hz$

  4. $1000 Hz$


Correct Option: A
Explanation:

The frequency (f) of a wave is the number of full wave forms 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, the time period is given as $1.18\times { 10 }^{ -3 }s$.
So, the frequency $f=\frac { 1 }{ t } =\frac { 1 }{ 0.00118s } =847.45Hz.$

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

  1. 0.02 s

  2. 0.04 s

  3. 0.05 s

  4. 0.08 s


Correct Option: A
Explanation:

The frequency (f) of a wave is the number of full wave forms 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, The piston in a petrol engine goes up and down 3000 times per minute, that is, 60 seconds. 
That is, the frequency is given as $f=\frac { 3000 }{ 60 } =50Hz$.
Time period = $\frac { 1 }{ f } =\frac { 1 }{ 50 } =0.02s$
Hence, the frequency is 50 Hz and the time period is 0.02 seconds.

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

  1. 15 Hz

  2. 10 Hz

  3. 40 Hz

  4. 5 Hz


Correct Option: D
Explanation:

The frequency (f) of a wave is the number of full wave forms 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 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.

Two trains A and B are approaching each other with $108 km {h}^{-1}$ and $126 kg {h}^{-1}$ respectively. If the train 'A' sounds a whistle of frequency 500 Hz, find the frequency of the whistle as heard by a passenger in the train 'B'.

(a) before the trains cross each other and

(b) after the trains cross each other. (Take velocity of sound as $330 {ms}^{-1}$)

  1. 608 Hz, 410 Hz

  2. 410 Hz, 608 Hz

  3. 330 Hz, 550 Hz

  4. 310 Hz, 660 Hz


Correct Option: A

In an open pipe pressure at the ends of the pipe is

  1. minimum

  2. maximum

  3. zero

  4. depending on temperature, it can be maximum or minimum.


Correct Option: A

If a vibrator strikes the water $50$ times in $5s$, then the frequency of wave is

  1. $10 Hz$

  2. $0.5 Hz$

  3. $5 Hz$

  4. $0.1 Hz$


Correct Option: A
Explanation:
Total number of vibrations, in $5s$ is $50$.
So, number of vibrations in $1s$ is $\dfrac{50}{5}=10$
Frequency$=10Hz$

A source of wave produce 1 crest and 1 trough in 16s. Find the frequency of the wave.

  1. .0625 Hz

  2. 16 Hz

  3. 100

  4. 0Hz


Correct Option: A
Explanation:

No of crest 1, trough = 1
$Time = 16 s$
No of wave in $1s = 1/16 = 0.0625 \ Hz$

A human heart on an average beat 72 times a minute. Calculate its frequency

  1. $72s^{-1}$

  2. $1.2s^{-1}$

  3. $4320s^{-1}$

  4. None


Correct Option: B
Explanation:

No of beat = $72 min^{-1}$ = $\dfrac {72}{1min} = \dfrac {72}{60}$=$1.2s^{-1}$
Frequency = $1.2s^{-1}$

One sound wave strikes a person's ear every $1/100$ of a second. The speed of sound where the person is located is $345 m/s$.
What is the frequency of the sound?

  1. $3.45\ Hz$

  2. $34,500\ Hz$

  3. $0.01\ Hz$

  4. $100\ Hz$

  5. $2.9\ Hz$


Correct Option: D
Explanation:

Time period of sound wave       $T = \dfrac{1}{100}$ second


$\therefore$ Frequency of sound wave      $\nu = \dfrac{1}{T} =100$  $Hz$

If you go on increasing the stretching force on a wire in a guitar, its frequency

  1. Increases

  2. Decreases

  3. Remains unchanged

  4. None of the above


Correct Option: A
Explanation:

When a stretching force is applied to a string or a wire, it applies a restoring force in the opposite direction. This force tends to bring the wire in its original length. It is known as tension in wire.

When we apply a stretching force on the wire of a guitar, the wire applies tension backwards. Greater the stretching force, greater is the tension.
The frequency of vibration of a stretched wire is directly proportional to the square root of the tension in it.
Hence on increasing the stretching force on a wire in a guitar, its frequency increases