Tag: superposition and interference of sound waves

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Multiple choice physics 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

  1. 1

  2. 2

  3. 4

  4. 8
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$y _1=A _0 \sin{\omega t}$, 

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

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

Multiple choice physics superposition and interference of sound waves

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

  1. $\dfrac{2\alpha}{1 + \alpha}$
  2. $\dfrac{2\sqrt{\alpha}}{1 + \alpha}$
  3. $\dfrac{2\alpha}{1 + \sqrt{\alpha}}$
  4. $\dfrac{1 + \alpha}{2\alpha}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)  } $

Multiple choice physics superposition and interference of sound waves

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


  1. a,d only

  2. b,d only

  3. a, c, d only

  4. a,b,c,d
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$\delta=\dfrac{\omega}{2\pi}=\dfrac{6\pi}{2\pi}=3Hz$

Multiple choice physics superposition and interference of sound waves

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:

  1. $75\,Hz$
  2. $150\,Hz$
  3. $200\,Hz$
  4. $300\,Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$75\ Hz, 225\ Hz, 375\ Hz$
Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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

  1. $10 s$
  2. $1 s$
  3. $0.1 s$
  4. $0.01 s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The beat frequency is the difference between the frequencies of two waves. The time interval between successive maximum intensities (beats) is the reciprocal of the beat frequency: T = 1/10 = 0.1 s.

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

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 

  1. $20$%
  2. $50$%
  3. $19$%
  4. $5$%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If intensity is reduced by 10 percent, 90 percent (0.9) remains. After two slabs, the remaining intensity is 0.9 * 0.9 = 0.81. The total reduction is 1 - 0.81 = 0.19, which is 19 percent.

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

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.

  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.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Intensity is proportional to the square of the amplitude and the square of the frequency (I proportional to A^2 * f^2). Wave 1 has amplitude 2a and frequency f, while Wave 2 has amplitude a and frequency 2f. Thus, intensities are proportional to (2a)^2 * f^2 = 4a^2f^2 and a^2 * (2f)^2 = 4a^2f^2, making them equal. Statement 2 is false because intensity depends on both amplitude and frequency.

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

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 :

  1. $25 : 1$
  2. $5 : 1$
  3. $9 : 4$
  4. $25 : 16$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } $

 

 

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

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.

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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