Tag: time period, frequency and amplitude of sound

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Questions Related to time period, frequency and amplitude of sound

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

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. $1000 Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $f=\frac { 1 }{ t } =\frac { 1 }{ 0.00118s } =847.45Hz.$

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Convert speeds to m/s: 108 km/h = 30 m/s, 126 km/h = 35 m/s. Before crossing, f' = f * ((v + vo) / (v - vs)) = 500 * ((330 + 35) / (330 - 30)) = 500 * (365 / 300) = 608.3 Hz. After crossing, f' = f * ((v - vo) / (v + vs)) = 500 * ((330 - 35) / (330 + 30)) = 500 * (295 / 360) = 409.7 Hz.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

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.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an open pipe, the ends are pressure nodes (displacement antinodes) because they are open to the atmosphere, meaning the pressure variation is zero relative to atmospheric pressure.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

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. $0.1 Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Total number of vibrations, in $5s$ is $50$.
So, number of vibrations in $1s$ is $\dfrac{50}{5}=10$
Frequency$=10Hz$
Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

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. $2.9\ Hz$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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


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

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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