Tag: interference of sound waves

Questions Related to interference of sound waves

Four independent waves are represented by the equations :
$y _1 = a _1\  sin\  \omega t$
$y _2 = a _2\ sin\  \omega t$
$y _3 = a _3\ cos\  \omega t$
$y _4 = a _4\ sin\  (\omega t + \pi/3)$ 
Then the waves for which phenomenon of interference will be observed are - 

  1. 1 and 3

  2. 1 and 4

  3. all 1, 2, 3 and 4

  4. None


Correct Option: A

Two sinusoidal plane waves same frequency having intensities $I _0 $ and $ 4I _0 $ are travelling in same direction. The resultant intensity at a point at which waves meet with a phase difference of zero radian is

  1. $ I _0$

  2. $5 I _0$

  3. $9 I _0$

  4. $3 I _0$


Correct Option: C
Explanation:

Let, $I _1=I _0  $ and $  I _2=4I _0 $

Resultant intensity, $I=I _1+I _2+2\sqrt{I _1I _2} cos\phi $
                                   $= I _0+4I _0+2\sqrt{I _04I _0} cos0^{\circ} \ =9I _0 $

If the ratio of maximum to minimum intensity in beat is 49, then the ratio of amplitudes of two progressive wave trains

  1. 7:1

  2. 4:3

  3. 49:1

  4. 16:9


Correct Option: B
Explanation:

$\dfrac{I _{max}}{I _{min}}=\dfrac{(\sqrt{I _1}+\sqrt{I _2})^2}{(\sqrt{I _1}-\sqrt{I _2})^2}=49$

$\dfrac{(\sqrt{I _1}+\sqrt{I _2})}{(\sqrt{I _1}-\sqrt{I _2})}=7$

$\sqrt{I _1}+\sqrt{I _2}=7(\sqrt{I _1}-\sqrt{I _2})$

$\dfrac{\sqrt{I _1}}{\sqrt{I _2}}=\dfrac{4}{3}$

$\dfrac{a}{b}=\dfrac{4}{3}$

Here, $a =\sqrt{I _1}$ and $b =\sqrt{I _2}$, where a and b are the amplitudes of the two progressive waves.

If the intensities of two interfering waves be $ I _1 $ and $ I _2  $, the contrast between maximum and minimum intensity is maximum, when

  1. $I _1 > > I _2$

  2. $I _1 < < I _2$

  3. $I _1 = I _2$

  4. either $I _1$ or $I _2$ is zero


Correct Option: C
Explanation:

$I _{max}=(\sqrt{I _1}+\sqrt{I _2})^2$
$I _{min}=(\sqrt{I _1}-\sqrt{I _2})^2$
Contrast is maximum, when $I _{min}=0$ ie. $I _1=I _2$

If the phase difference between two sound waves of wavelength $  \lambda  $ is $  60^{\circ} $, the corresponding path difference is

  1. $ \frac{\lambda}{6} $

  2. $ \frac{\lambda}{2} $

  3. $ \lambda 2 $

  4. $ \frac{\lambda}{4} $


Correct Option: A

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 

  1. 6

  2. 5

  3. 4

  4. 2


Correct Option: A

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

  1. 9 and 1

  2. 31 and 1

  3. 91 and 31

  4. 61 and 1


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 :

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


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.

  1. True

  2. False


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 

  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 $n=0,1,2$


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, ..... $