Tag: physics

Questions Related to physics

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

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

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. $3 m/sec$

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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
$75\ Hz, 225\ Hz, 375\ Hz$

Which of the following function represent traveling waves?

standing waves waves physics

  1. $ y = (x + 5t)^3 $

  2. $ y = tan (2x + 3t) $

  3. $ y = \theta^{(4t+2x)^2} $

  4. $ y = \frac { 1 }{ x + 3t } $

Show answer
Correct Option: D