Tag: sound: production of sound

Questions Related to sound: production of sound

Two sound waves having pressure
$P _{1}=2 \times 10^{4} \sin (2\pi \times 10^{4}\ t)Pa$ and 
$P _{2}=4 \times 10^{4} \sin (3\pi \times 10^{4}\ t+\pi /6)Pa$
superimpose with each other. Find the amplitude of resultant wave.

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

  1. $4.47\times 10^{4}\ Pa$

  2. $4.47\ Pa$

  3. $5.67\times 10^{4}\ Pa$

  4. $5.67\ Pa$

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Correct Option: A

Two tuning forks of frequency 256 and 258 vibrating per second are sounded together, then time interval between consecutive maxima heard by the observer is.

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

  1. 2 sec

  2. 1 sec

  3. 0.5 sec

  4. 1/4 sec

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Correct Option: C
Explanation:

$\begin{array}{l} { n _{ 1 } }=256 \ { n _{ 2 } }=258 \ Both\, tuning\, forks\, are\, making\, beats \ so,\, we\, know\, that \ { t _{ \left( { for\, \, \max  . } \right)  } }=\dfrac { 1 }{ { { n _{ 2 } }-{ n _{ 1 } } } } \sec   \ =\dfrac { 1 }{ { 258-256 } }  \ =0.5\, \sec   \end{array}$

Hence, the option $C$ is the correct answer.

The frequency of light of wave length $5000{A^\circ}$ is :

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

  1. $1.5 \times 10^3 Hz$

  2. $6 \times 10^8 Hz$

  3. $6 \times 10^{14} Hz$

  4. $7.5 \times 10^{15} Hz$

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Correct Option: C
Explanation:

We know that $ n = \dfrac{v}{\lambda } $
where, $ v = $ velocity of light
$\lambda  =$ wavelenght
So, $ n = \dfrac{ 3 \times  10^{8}}{5000 \times  10^{-10} } = 6 \times 10^{14} Hz$

The relation between frequency(f) and time period(T) is given by

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

  1. f = T

  2. f = $\frac{1}{T}$

  3. T = f$^2$

  4. T = f$^3$

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Correct Option: B
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 time period are reciprocals. 
That is, T=1/f.  

If the frequency of sound wave in air is doubled, its wavelength.

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

  1. is doubled

  2. increases 4 times

  3. decreases 4 times

  4. is halved

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Correct Option: D
Explanation:

Answer is D.

A sound wave has a speed that is mathematically related to the frequency and the wavelength of the wave. The mathematical relationship between speed, frequency and wavelength is given by the following equation.
Speed = Wavelength * Frequency
That is, Frequency = speed / Wavelength. The frequency and wavelength are inversely proportional to each other.
So, when the frequency is doubled, the wavelength will be halved.

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

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

  1. $847.45 Hz$

  2. $800 Hz$

  3. $935. 55 Hz$

  4. $1000 Hz$

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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.

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

  1. 0.02 s

  2. 0.04 s

  3. 0.05 s

  4. 0.08 s

Show answer
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.

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

  1. 15 Hz

  2. 10 Hz

  3. 40 Hz

  4. 5 Hz

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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 $330 {ms}^{-1}$)

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

  1. 608 Hz, 410 Hz

  2. 410 Hz, 608 Hz

  3. 330 Hz, 550 Hz

  4. 310 Hz, 660 Hz

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Correct Option: A

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

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

  1. minimum

  2. maximum

  3. zero

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

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Correct Option: A