Tag: superposition and interference of sound waves

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Questions Related to superposition and interference of sound waves

Multiple choice physics superposition and interference of sound waves

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

  1. $I = 100\ I _{0}$
  2. $I = 10\ I _{0}$
  3. $I = I _{0}$
  4. $I = \surd {10}\ I _{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For incoherent sources (random phases), the intensities add linearly. If there are 10 sources each with intensity I0, the total intensity is 10 * I0.

Multiple choice physics superposition and interference of sound waves

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 $350m/sec$)

  1. $0.7mm$
  2. $0.1mm$
  3. $0.2mm$
  4. $0.5mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice physics superposition and interference of sound waves

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?

  1. $8$
  2. $6$
  3. $4$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The beat frequency is the difference between the two frequencies. f1 = 100/2 = 50 Hz, f2 = 92/2 = 46 Hz. Beat frequency = 50 - 46 = 4 Hz.

Multiple choice physics superposition and interference of sound waves

 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: 

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

The resultant amplitude of two waves with amplitude A and phase difference phi is sqrt(A^2 + A^2 + 2*A*A*cos(phi)). For phi = pi/2, this is sqrt(2*A^2) = A*sqrt(2). The frequency remains the same.

Multiple choice physics superposition and interference of sound waves

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

  1. 4 : 1

  2. 9 : 1

  3. 2 : 1

  4. 10 : 8
Reveal answer Fill a bubble to check yourself
A Correct answer
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       $\dfrac{I _{max}}{I _{min}}  =  \dfrac{(\sqrt{I _1} + \sqrt{I _2})^2}{(\sqrt{I _2} - \sqrt{I _1})^2} = \dfrac{(\sqrt{I _1} + \sqrt{9  I _1})^2}{(\sqrt{9  I _1} - \sqrt{I _1})^2} = \dfrac{16  I _1}{4  I _1}$ 

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

Multiple choice physics superposition and interference of sound waves

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

  1. $a^{2}cos^{2}\delta /2$
  2. $2a^{2}cos^{2}\delta /2$
  3. $3a^{2}cos^{2}\delta /2$
  4. $4a^{2}cos^{2}\delta /2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The resultant intensity of two interfering waves is proportional to the square of the resultant amplitude. Using phasor addition for waves with identical amplitude a and phase difference delta, the resultant amplitude is 2a cos(delta/2). Squaring this gives an intensity proportional to 4a^2 cos^2(delta/2).

Multiple choice physics superposition and interference of sound waves

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 $n$?

  1. $4$
  2. $2$
  3. $\sqrt2$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $A _0$.

Multiple choice physics superposition and interference of sound waves

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

  1. $5$
  2. $5\sqrt{2}$
  3. $5\sqrt{3}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice physics superposition and interference of sound waves

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 

  1. $\displaystyle I _0$
  2. $\displaystyle 2 I _0$
  3. $\displaystyle 4I _0$

  4. Zero
Reveal answer Fill a bubble to check yourself
B Correct answer
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       $< I > = 4I _o  <cos^2  \dfrac{\delta}{2}>$

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