Tag: sound: production of sound

Quizzes (4) Questions (118)

Questions Related to sound: production 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

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.

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

  2. $4.47\ Pa$

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

  4. $5.67\ Pa$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The resultant amplitude of two waves with different frequencies is not a simple sum. However, if interpreting as phasors or peak pressure values, the maximum resultant amplitude is sqrt(P1^2 + P2^2 + 2*P1*P2*cos(phi)). With P1=2e4, P2=4e4, and phase difference, the calculation yields approximately 4.47e4 Pa.

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

  1. 2 sec

  2. 1 sec

  3. 0.5 sec

  4. 1/4 sec
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

  1. $1.5 \times 10^3 Hz$

  2. $6 \times 10^8 Hz$

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

  4. $7.5 \times 10^{15} Hz$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$

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

  1. f = T

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

  3. T = f$^2$

  4. T = f$^3$
Reveal answer Fill a bubble to check yourself
B 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 time period are reciprocals. 
That is, T=1/f.  

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

  1. is doubled

  2. increases 4 times

  3. decreases 4 times

  4. is halved
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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.