Tag: superposition and interference of sound waves

Questions Related to superposition and interference of sound waves

Ration of maximum and minimum intensities is refrence pattern is 25:1 . The ration of intensities of refring waves is:

  1. 25 : 1

  2. 5 : 1

  3. 6 : 4

  4. 625 : 1


Correct Option: C

Two waves of intensities $I$ and $4I$ superimpose. The minimum and maximum intensities will respectively be

  1. $I,\space 9I$

  2. $3I,\space 5I$

  3. $I,\space 5I$

  4. None of these


Correct Option: A
Explanation:
The Intensity of the wave is directly proportional to the square of its amplitude.
$I \propto A^{2}$;
$I = cA$$^2$;   '$c$' is an arbitrary constant.
So, if a wave with Intensity $I$ has an amplitude of $A(A{ _{1}}$)
A wave with an Intensity of $4I$ would respectively have an amplitude of $2A$($A{ _{2}}$)

If 2 waves with amplitudes $A{ _{1}}$,$A{ _{2}}$ are superimposed, the resultants would be
Maximum of $A{ _{1}}$ + $A{ _{2}} = 3A$ (Constructive Interference)
Minimum of $|A{ _{1}}$ - $A{ _{2}} |   = A $ (Destructive Interference)
The wave of amplitude $3A$ would have an Intensity of $9I$
The wave of amplitude $A$ would have an Intensity of $I$

For a wave displacement amplitude is $10^{-8} m$ density of air $1.3 kg m^{-3}$ velocity in air $340 ms^{-1}$ and frequency is 2000 Hz.The average intensity of wave is

  1. $5.3\times 10^{-4} Wm^{-2}$

  2. $5.3\times 10^{-6} Wm^{-2}$

  3. $3.5\times 10^{-8} Wm^{-2}$

  4. $3.5\times 10^{-6} Wm^{-2}$


Correct Option: A

Two sound waves of equal intensity $I$ superimpose at point $P$ in $90^{\small\circ}$ out of phase. The resultant intensity at point $P$ will be

  1. $4I$

  2. $\sqrt2I$

  3. $2I$

  4. $I$


Correct Option: C
Explanation:

Amplitude of the resultant wave is:
$A _R=\sqrt{A^2 _1+A^2 _2+2A _1A _2\cos(\theta)}$.

Here, $A _1=A _2=A \text{ and } \theta = \pi/2$

So, $A _R=A\sqrt{2(1+\cos(\theta))}=2A\cos(\theta/2)=2A\cos(\pi/4)=\sqrt{2}A$ 

$\Rightarrow I _R=|A _R|^2=2A^2=2I$

A wave of frequency 500$\mathrm { Hz }$ travels between $\mathrm { x }$and $\mathrm { Y }$ and travel a distance of 600$\mathrm { m }$ in 2$\mathrm { sec }$ . between $X$ and $Y .$ How many wavelength are therein distance $X Y$ :

  1. 1000

  2. 300

  3. 180

  4. 2000


Correct Option: A

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