Tag: superposition and interference of sound waves

Questions Related to superposition and interference of sound waves

Equations of stationary and a travelling wave are as follows: $Y _1=sin\, kx\, cos\,\omega t$ and $Y _2=a\, sin\, (\omega t-kx)$. The phase difference between two points $X _1=\dfrac{\pi}{3k}$ and $ X _2=\dfrac{3\pi}{2k}$ are $\phi _1$ and $\phi _2$ respectively for the two waves.The ratio of $\dfrac{\phi _1}{\phi _2}$ is 

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

  1. 6

  2. 5

  3. 4

  4. 2

Show answer
Correct Option: A

Two waves of intensities 1 and 4 superimposes. Then the maximum and minimum intensities are :

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

  1. 9 and 1

  2. 31 and 1

  3. 91 and 31

  4. 61 and 1

Show answer
Correct Option: A
Explanation:

Ratio of amplitudes $\sqrt{\dfrac{4}{1}}=\dfrac{2}{1}$
$\dfrac{maximum\ amplitude}{minimum\ amplitude}=\dfrac{2+1}{2-1}=\dfrac{3}{1}$


$\dfrac{maximum\ intensity}{minimum\ intensity}=\left (\dfrac{3}{1}  \right )^2=\dfrac{9}{1}$

Two periodic waves of intensities ${I} _{1}$ and ${I} _{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities possible is :

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

  1. ${I} _{1} + {I} _{2}$

  2. ${\left(\sqrt{{I} _{1}} + \sqrt{{I} _{2}}\right)}^{2}$

  3. ${\left(\sqrt{{I} _{1}} - \sqrt{{I} _{2}}\right)}^{2}$

  4. $2\left({I} _{1} + {I} _{2}\right)$

Show answer
Correct Option: D
Explanation:

Resultant intensity of two periodic waves is given by
$I={ I } _{ 1 }+{ I } _{ 2 }+2\sqrt { { I } _{ 1 }{ I } _{ 2 }\cos { \delta  }  } $
where $\delta$ is the phase difference between the waves.
For maximum intensity,
$\delta =2n\pi ;n=0,1,2,...$etc.
Therefore, for zero order maxima, $\cos { \delta  } =1$
${ I } _{ max }={ I } _{ 1 }+{ I } _{ 2 }+2\sqrt { { I } _{ 1 }{ I } _{ 2 } } ={ \left( \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
For minimum intensity,
$\delta =\left( 2n-1 \right) \pi ;n=1,2,...$etc.
Therefore, for Ist order minima, $\cos { \delta  } =-1$
${ I } _{ min }={ I } _{ 1 }+{ I } _{ 2 }-2\sqrt { { I } _{ 1 }{ I } _{ 2 } } $
$={ \left( \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
Therefore,  ${ I } _{ max }+{ I } _{ min }={ \left( \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }+{ \left( \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
$=2\left( { I } _{ 1 }+{ I } _{ 2 } \right) $

State whether true or false :
The phenomenon of interference is consistent with the law of conservation of momentum.

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The correct answer is option(B).

Interference is nothing but simply a phenomena of redistribution of energy. If it is constructive then energy is increased and hence intensity also and if it is destructive then energy is decreased and hence intensity also. So energy is redistributed in constructive and destructive interference but remains conserved.

A travelling wave represented by $y=A\sin { \left( \omega t-kx \right)  } $ is superimposed on another wave represented by $y=A\sin { \left( \omega t+kx \right)  } $. The resultant is 

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

  1. A standing wave having nodes at$\quad x=\left( n+\cfrac { 1 }{ 2 } \right) \cfrac { \lambda }{ 2 } $, where $n=0,1,2$

  2. A wave travelling along $+x$ direction

  3. /a wavelength travelling along $-x$ direction

  4. a standing wave having nodes at $x=\cfrac { n\lambda }{ 2 } $, where $n=0,1,2$

Show answer
Correct Option: A
Explanation:

According to the principle of superposition, the resultant wave is
$y = asin(kx - \omega t) + asin(kx + \omega t)$
$= 2a\ sin\ \omega t\ cos\ x$                                                 .....(i)

It represents a standing wave.
In the standing wave, there will be nodes (where amplitude is zero) and antinodes  (where amplitude is largest).
From Eq. (i), the positions of nodes are given by
$sin\ kx = 0 \Longrightarrow kx = n\pi; n = 0, 1, 2, ....$

or $\dfrac{2\pi}{\lambda}x = n\pi; 0, 1, 2, ....$

or $x = \dfrac{n\lambda}{2}; n = 0, 1, 2, ...$
In the same way,
From Eq.(i), the positions of antinodes are given by$|sinkx| = 1$
$\Longrightarrow kx = (n + \dfrac{1}{2})\pi ; n = 0, 1, 2, ..... $

or $\dfrac{2\pi x}{\lambda} =  (n + \dfrac{1}{2})\pi ; n = 0, 1, 2, ..... $

or $x =  (n + \dfrac{1}{2})\dfrac{\lambda}{2} ; n = 0, 1, 2, ..... $

The ratio of intensities of two waves that produce interference pattern is 16:1, then the ratio of maximum and minimum intensities in the pattern is :

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

  1. 25:9

  2. 9:25

  3. 1: 4

  4. 4:1

Show answer
Correct Option: A
Explanation:

Let the intensities of the two waves be $I _1$ and $I _2$.
Given :  $I _1:I _2 = 16:1$
Ratio of maximum and minimum intensities  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{I _1}  +\sqrt{I _2}}{\sqrt{I _1} - \sqrt{I _2}}\bigg)^2$
Or   $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{\frac{I _1}{I _2}}  +1}{\sqrt{\frac{I _1}{I _2}} - 1}\bigg)^2$

Or  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{16}  +1}{\sqrt{16} - 1}\bigg)^2 = \bigg(\dfrac{4+1}{4-1}\bigg)^2$
$\implies  \ $  $\dfrac{I _{max}}{I _{min}} = \dfrac{25}{9}$

Consider the superposition of N harmonic waves of equal amplitude and frequency. If N is a very large number determine the resultant intensity in terms of the intensity $\left( { I } _{ 0 } \right)$ of each component wave for the condition when the component waves have identical phases.

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

  1. ${ NI } _{ 0 }$

  2. ${ N }^{ 2 }{ I } _{ 0 }$

  3. $\sqrt { N } { I } _{ 0 }$

  4. ${ I } _{ 0 }$

Show answer
Correct Option: A
Explanation:

As all the waves are in phase and having same amplitude and frequency

So, the intensities will simply get add to give the resultant intensity
$\Rightarrow (Intensity) _{Resultant}=(I _0+I _0+.....I _0)-N\quad times\ \quad\quad\quad\quad\quad=NI _0$

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

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

  1. 2:1

  2. 9:1

  3. 3:1

  4. 4:1

Show answer
Correct Option: D
Explanation:

Let the intensities of the two waves be $I _1$ and $I _2$.
Given :  $I _1:I _2 = 9:1$
Ratio of maximum and minimum intensities  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{I _1}  +\sqrt{I _2}}{\sqrt{I _1} - \sqrt{I _2}}\bigg)^2$
Or   $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{\frac{I _1}{I _2}}  +1}{\sqrt{\frac{I _1}{I _2}} - 1}\bigg)^2$

Or  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{9}  +1}{\sqrt{9} - 1}\bigg)^2 = \bigg(\dfrac{3+1}{3-1}\bigg)^2$
$\implies  \ $  $\dfrac{I _{max}}{I _{min}} = \dfrac{16}{4} = \dfrac{4}{1}$

Assertion - Two sinusoidal waves on the same string exhibit interference.
Reason - these waves, add or cancel out according to the principle of superposition

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

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Both Assertion and Reason are incorrect

Show answer
Correct Option: A
Explanation:

When there are two sinusoidal waves in a string, they cause interference of the waves. The principle of superposition is basic to the phenomenon of interference.

So, the assertion and reason both are correct and the reason is the correct explanation for the assertion.

Which of the following is true?

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

  1. Both the light and sound waves exhibit interference

  2. Light waves exhibit interference

  3. Sound waves exhibit interference

  4. Neither sound waves nor light waves exhibit interference

Show answer
Correct Option: A