Tag: superposition and interference of sound waves

Questions Related to superposition and interference of sound waves

Four sources of sound each of sound level 10 dB are sounded together in phase, the resultant intensity level will be ($log _{10}2 = 0.3$)

physics superposition and interference of sound waves

  1. 40 dB

  2. 26 dB

  3. 22 dB

  4. 13 dB

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Correct Option: D

Consider ten identical sources of sound all giving the same frequency but having phase angles which are random. If the average intensity of each source is $I _{0}$, the average of resultant intensity $I$ due to all these ten sources will be

physics superposition and interference of sound waves

  1. $I = 100\ I _{0}$

  2. $I = 10\ I _{0}$

  3. $I = I _{0}$

  4. $I = \surd {10}\ I _{0}$

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Correct Option: B

Two sources of sound A and B produce the wave of $350Hz$, they vibrate in the same phase. The particle $P$ is vibrating under the influence of these two waves. If the amplitude at the point $P$ produced by the two waves is $0.3mm$ and $0.4mm$ then the resultant amplitude of the point $P$ will be: (path difference $AP-BP=25cm$ and the velocity of sound is $350m/sec$)

physics superposition and interference of sound waves

  1. $0.7mm$

  2. $0.1mm$

  3. $0.2mm$

  4. $0.5mm$

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Correct Option: D
Explanation:

$\begin{array}{l} \lambda =\dfrac { v }{ n } =\dfrac { { 350 } }{ { 350 } } =1m=100cm \ Path\, difference\, \, \Delta x=AP-BP=25cm \ Phase\, difference\, \, \Delta \varphi =\dfrac { { 2\pi  } }{ \lambda  } \Delta x=\dfrac { { 2\pi  } }{ 1 } \times \left( { \dfrac { { 25 } }{ { 100 } }  } \right) =\dfrac { \pi  }{ 2 }  \ Amplitude\, A=\sqrt { { { \left( { { a _{ 1 } } } \right)  }^{ 2 } }+{ { \left( { { a _{ 2 } } } \right)  }^{ 2 } } } =\sqrt { { { \left( { 0.3 } \right)  }^{ 2 } }+{ { \left( { 0.4 } \right)  }^{ 2 } } } =0.5mm \end{array}$

Two plane harmonic sound waves travelling in the same direction are given by the following displacement equations
$y _{1} (x, t) = A\cos (0.5\pi x - 100\pi t)$
$y _{2} (x, 1) = A\cos (0.46\pi x - 92\pi t)$
How may times, a listener can hear sound of maximum intensity in one second?

physics superposition and interference of sound waves

  1. $8$

  2. $6$

  3. $4$

  4. $3$

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Correct Option: C

 When two sound waves with a phase of $\dfrac { \pi  }{ 2 } $ and each having amplitude A and frequency $\omega $, are superimposed on each other, then the maximum amplitude and frequency  of resultant wave is: 

physics superposition and interference of sound waves

  1. $\sqrt { 2 } A;\omega $

  2. $\dfrac { A }{ \sqrt { 2 } } ;\dfrac { \omega }{ 2 } $

  3. $\left( \sqrt { 2 } \right) A;\dfrac { \omega }{ 2 } $

  4. $\dfrac { A }{ \sqrt { 2 } } ;\omega $

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Correct Option: A

Two waves having the intensities in the ratio 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to

physics superposition and interference of sound waves

  1. 4 : 1

  2. 9 : 1

  3. 2 : 1

  4. 10 : 8

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Correct Option: A
Explanation:

Let the intensity of the two waves be $I _1$  and  $I _2$


Given:    $I _2  :  I _1  =  9  :  1        \implies  I _2   =  9  I _1$

Now       $\dfrac{I _{max}}{I _{min}}  =  \dfrac{(\sqrt{I _1} + \sqrt{I _2})^2}{(\sqrt{I _2} - \sqrt{I _1})^2} = \dfrac{(\sqrt{I _1} + \sqrt{9  I _1})^2}{(\sqrt{9  I _1} - \sqrt{I _1})^2} = \dfrac{16  I _1}{4  I _1}$ 

$\implies    I _{max}  :  I _{min}  =  4  : 1$

Two waves $Y _{1}= asin\omega t$  and $Y _{2}= asin(\omega t+\delta )$  are  producing interference, then resultent intensity is 

physics superposition and interference of sound waves

  1. $a^{2}cos^{2}\delta /2$

  2. $2a^{2}cos^{2}\delta /2$

  3. $3a^{2}cos^{2}\delta /2$

  4. $4a^{2}cos^{2}\delta /2$

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Correct Option: A

Beats are produced because of the superposition of two progressive notes> Maximum loudness at the waxing is $n$ times the loudness of either notes. What is the values of $n$?

physics superposition and interference of sound waves

  1. $4$

  2. $2$

  3. $\sqrt2$

  4. $1$

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Correct Option: A
Explanation:

Resultant amplitude is $A=\sqrt{A^2 _1+A^2 _2+2A _1A _2\cos(\phi)}$, where 


$\phi$ is the angle between the superposing waves.

$A=A _{max}=A _1+A _2$, when $\cos(\phi)=1$.

So, maximum loudness $I _{max}=A^2 _{max}=(A _1+A _2)^2=4A^2 _0=4I _0$, 

assuming that the waves have same amplitude $A _0$.

If a tuning fork sends a wave $5 sin \displaystyle \left(600\omega t - \frac{\pi}{0.6}x \right)$, then the amplitude of the intensity heard is

physics superposition and interference of sound waves

  1. $5$

  2. $5\sqrt{2}$

  3. $5\sqrt{3}$

  4. none of these

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Correct Option: A
Explanation:

$\displaystyle \Delta x = 0.4m$
 $\displaystyle\Phi = k\Delta x = \frac{1\pi}{3} = \sqrt{5^2+5^2+2\times 5 cos  (2\pi/3)}$
$\displaystyle= 5.$

Two identical sources of sound of same frequency and identical intensities $\displaystyle I _0$ are producing sound. If their phases are irregular, then the average intensity of sound at a point where waves from the two sources are superposing is 

physics superposition and interference of sound waves

  1. $\displaystyle I _0$

  2. $\displaystyle 2 I _0$

  3. $\displaystyle 4I _0$

  4. Zero

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Correct Option: B
Explanation:

Given :   $I _1 = I _2  = I _o$


Resultant intensity        $I = I _1 + I _2 + 2  \sqrt{I _1  I _2}   cos \delta   $         
where $\delta $ is the phase difference.

 $I = I _o + I _o + 2  \sqrt{I _o \times  I _o}   cos \delta   = 2 (1 + cos  \delta)   $  

$\implies  I = 4  I _o  cos^2 \dfrac{\delta}{2}$

Now average intensity       $< I > = 4I _o  <cos^2  \dfrac{\delta}{2}>$

 $< I > = 4I _o  \times \dfrac{1}{2}  =  2  I _o$