Tag: superposition and interference of sound waves

Questions Related to superposition and interference of sound waves

When interference is produced by two progressive waves of equal frequencies, then the maximum intensity of the resulting sound are N times the intensity of each of the component waves. The value of N is

physics superposition and interference of sound waves

  1. 1

  2. 2

  3. 4

  4. 8

Show answer
Correct Option: C
Explanation:

$y _1=A _0 \sin{\omega t}$, 

$y _2=A _0 \sin{\omega t+ \phi}$,

also, $I \propto y^2={(y _1+y _2)}^2={(2A _0 \sin{\omega t +\phi/2} \cos{\phi/2})}^2=4{A _0}^2{(\sin{\omega t +\phi/2})}^2{(\cos{\phi/2})}^2$,

Two coherent sources of intensity ratio $\alpha$ interfere. In interference pattern $\dfrac{{I} _{max} - {I} _{min}}{{I} _{max} + {I} _{min}} =$

physics superposition and interference of sound waves

  1. $\dfrac{2\alpha}{1 + \alpha}$

  2. $\dfrac{2\sqrt{\alpha}}{1 + \alpha}$

  3. $\dfrac{2\alpha}{1 + \sqrt{\alpha}}$

  4. $\dfrac{1 + \alpha}{2\alpha}$

Show answer
Correct Option: B
Explanation:

$\dfrac { { I } _{ max }-{ I } _{ min } }{ { I } _{ max }+{ I } _{ min } } =\dfrac { { \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 }-{ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 } }{ { \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 }+{ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 } }$
                                       [$\because { I } _{ max }={ \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 },{ I } _{ min }={ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 }$  where $a =$ amplitude]
                $=\dfrac { 4{ a } _{ 1 }{ a } _{ 2 } }{ 2\left( { a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 } \right)  } =\dfrac { 2{ a } _{ 1 }{ a } _{ 2 } }{ { a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 } } $
Now, dividing the numerator and denominator by ${a} _{1}{a} _{2}$, we get
$\dfrac { { I } _{ max }-{ I } _{ min } }{ { I } _{ max }+{ I } _{ min } } =\dfrac { 2 }{ \left[ \dfrac { { a } _{ 1 } }{ { a } _{ 2 } } +\dfrac { { a } _{ 2 } }{ { a } _{ 1 } }  \right]  } =\dfrac { 2 }{ \left[ \sqrt { \alpha  } +\dfrac { 1 }{ \sqrt { \alpha  }  }  \right]  } =\dfrac { 2\sqrt { \alpha  }  }{ \left( \alpha +1 \right)  } $

In case of super position of waves (at $x=0$),
 $y _{1}=4\sin(1026\pi t)$ and $y _{2}=2\sin(1014\pi t)$


a) the frequency of resulting wave is $510$ Hz
b) the amplitude of resulting wave varies at the frequency of $3$ Hz
c) the frequency of beats is $6$ per second
d) the ratio of maximum to minimum intensity is $9$

The correct statements are


physics superposition and interference of sound waves

  1. a,d only

  2. b,d only

  3. a, c, d only

  4. a,b,c,d

Show answer
Correct Option: D
Explanation:

Beat $=\delta _{1}-\delta _{2}$$=\dfrac{\omega _{1}}{2\pi}-\dfrac{\omega _{2}}{2\pi}$$=\dfrac{1026\pi}{2\pi}-\dfrac{1014\pi}{2\pi}$$=6$

$\dfrac{I max}{I min}=\dfrac{(\Delta _{1}+A _{2})^{2}}{(\Delta _{1}A _{2})^{2}}=\dfrac{(4+2)^{2}}{(4-2)^{2}}=\dfrac{36}{4}=\dfrac{9}{1}$

$y _{1}=4 sin (1026 \pi t)$
$y _{2}=2 sin (1014 \pi t)$
$y=y _{1}+y _{2}$
   $=4 Sin (1026 \pi t)+2 sin (1014\pi t)$
   $=\left(4\sqrt{(\dfrac{3}{1})^{2}+cos (12\pi t)}\right ) sin (1020 \pi t)$

So, clearly frequency $ =\dfrac{\omega}{2\pi}=\dfrac{1020\pi}{2\pi}=510 Hz$
and amplitude of resulting wave varies at frequency
$\delta=\dfrac{\omega}{2\pi}=\dfrac{6\pi}{2\pi}=3Hz$

If an observer is walking away from the plane mirror with $6 m/sec$. Then the velocity of the image with respect to the observer will be

physics superposition and interference of sound waves

  1. $6 m/sec$

  2. $-6 m/sec$

  3. $12 m/sec$

  4. $3 m/sec$

Show answer
Correct Option: D

A loudspeaker that produces signals from $50Hz$ to $500Hz$ is placed at the open end of a closed tube of length $1.1m.$ If velocity of sound is $330m/s,$ then frequencies that excites resonance in the tube are:

physics superposition and interference of sound waves

  1. $75\,Hz$

  2. $150\,Hz$

  3. $200\,Hz$

  4. $300\,Hz$

Show answer
Correct Option: A
Explanation:
If the length of the tube is $l$
then $l=\dfrac{x}{4}\Rightarrow \lambda =4l$
where $\lambda$ is the wavelength of the sound wave inside the tube If corresponding frequency is $V$ than 
$V\lambda=V$
$\therefore V=\dfrac{V}{\lambda}=\dfrac{V}{4l}=\dfrac{330}{4.(1.1)}$
$\therefore V=75\ Hz$
Therefore the fundamental tone freq of the tube is $V=75\ Hz$
This frequencies of the overtones are $3v, 5v, 3v .... $
i.e. $225, 373, 525$
If the loud speaker produces signals from $0\ Hz$ to $500\ Hz$ then frequencies that excites resonance in the tube are
$75\ Hz, 225\ Hz, 375\ Hz$

If the difference between the frequencies of two waves is 10 Hz then time interval between successive maximum intensity is:

physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

  1. $10 s$

  2. $1 s$

  3. $0.1 s$

  4. $0.01 s$

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Correct Option: C

The intensity of the sound gets reduced by $10$% on passing through a slab. The reduction in  intensity on passing through two consecutive slab, would be 

physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

  1. $20$%

  2. $50$%

  3. $19$%

  4. $5$%

Show answer
Correct Option: C

Statement-1:
Two longitudinal waves given by equations; ${ y } _{ 1 }$(x,t) = 2a $\sin { \left( \omega t-kx \right)  } $ and ${ y } _{ 2 }\left( x,t \right) $ = a $\sin { \left( 2\omega t-2kx \right)  } $  will have equal intensity.

Statement-2:
Intensity of waves of given frequency in same medium is proportional to square of amplitude only.

physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

  1. Statement-1 is true, statement -2 is true; statement-2 is not correct explanation of statement -1.

  2. Statement-1 is false, statement -2 is true.

  3. Statement-1 is true, statement -2 is false.

  4. Statement -1 is true, statement-2 true; statement-2 is the correct explanation of statement-1.

Show answer
Correct Option: C

Two coherent sources of different intensities send waves which interfere. If the ratio of maximum and minimum intensity in the interference pattern is $25$ then find ratio of intensity of source :

physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

  1. $25 : 1$

  2. $5 : 1$

  3. $9 : 4$

  4. $25 : 16$

Show answer
Correct Option: C
Explanation:

$\cfrac { { I } _{ max } }{ { I } _{ min } } ={ \left[ \cfrac { \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  }{ \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  }  \right]  }^{ 2 }$

Where and  are intensities of two waves 

given 

$\cfrac { { I } _{ max } }{ { I } _{ min } } =\cfrac { 25 }{ 1 } \\ \therefore \cfrac { \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  }{ \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  } =\cfrac { 5 }{ 1 } $

use componendo and dividendo

we get

$\cfrac { \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } } +\sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  }{ \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } } -\sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  } =\cfrac { 5+1 }{ 5-1 } \\ or,\quad \cfrac { \sqrt { { I } _{ 1 } }  }{ \sqrt { { I } _{ 2 } }  } =\cfrac { 3 }{ 2 } \\ or,\quad \cfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\cfrac { 9 }{ 4 } $

 

 

Statement -1:
Two longitudinal waves given by equation $y _{1}$(x,t) = 2a sin $(\omega - kx)$ and $y _{2}$(x,t) = a sin $(2\omega - 2kx)$ will have equal intensity.
Statement -2:
Intensity of waves of given frequency in the same medium is proportional to the square of amplitude only.

physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

  1. Statement -1 is true, statement -2 is true; statement -2 is not correct explanation of statement-1.

  2. Statement -1 is false, statement -2 is true.

  3. Statement -1 is true, statement -2 is false

  4. Statement -1 is true, statement -2 true; statement -2 is the correct explanation of statement -1

Show answer
Correct Option: C
Explanation:

Intensity of waves of given frequency in same medium is not only proportional to square of amplitude. It depends on other factors also