Tag: properties of sound waves

Questions Related to properties of sound waves

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

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

  1. $10 s$

  2. $1 s$

  3. $0.1 s$

  4. $0.01 s$

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Correct Option: C

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 

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

  1. $20$%

  2. $50$%

  3. $19$%

  4. $5$%

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Correct Option: C

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.

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

  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.

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Correct Option: C

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 :

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

  1. $25 : 1$

  2. $5 : 1$

  3. $9 : 4$

  4. $25 : 16$

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Correct Option: C
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 } $

 

 

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.

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

  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

Show answer
Correct Option: C
Explanation:

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

If two waves maintain constant phase difference or same phase at any two points on a wave is known as spatial coherence.

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

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Correct Option: A
Explanation:

Two wave sources are said to be coherent if the waves either have same phase or constant phase difference at any two points on a wave and the phenomenon is known as spatial coherence.

Intensity (I) is related to amplitude (A) as:

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

  1. $I \propto A$

  2. $I \propto {A^2}$

  3. $I \propto {A^4}$

  4. $I \propto {A^{-1}}$

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Correct Option: B
Explanation:

Intensity of a wave is directly proportional to the square of amplitude of the wave.
i.e.  $I \propto A^2$
More is the amplitude of wave, more will be the intensity.
So, we write  $I = kA^2$
where $k$ is a proportionality constant.

The resultant intensity for two identical waves of intensity I with a phase difference of $\pi/3$ is 

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

  1. $R=2\sqrt{3I}$

  2. $R=\sqrt{3I}$

  3. $R=4\sqrt{3I}$

  4. $R=3\sqrt{3I}$

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Correct Option: B
Explanation:

The resultant of two waves is given by $R^2=A^2+B^2+2AB cos \theta; \theta$ is the phase difference and A and B are the amplitudes of the individual waves.

Since both the waves are identical, their amplitudes are equal

Thus, $R^2 = A^2+A^2+A^2=3A^2 \implies R=\sqrt{3}A$

Since intensity is proportional to $\sqrt{A}$, we can write the resultant amplitude $R = \sqrt{3I}$

The correct option is (b)

If the sum of the intensities of two component waves are 5I units and upon superposition with a phase difference of $\pi $ radians, their resultant is 2I, what are the intensities of component waves

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

  1. $I _1=3I, I _2=I$

  2. $I _1=3.5I, I _2=1.5I$

  3. $I _1=2I, I _2=3I$

  4. $I _1=I, I _2=4I$

Show answer
Correct Option: B
Explanation:

If $I _1$ and $I _2$ are intensities of two waves, then $I _1+I _2=5I$ and $(I _1-I _2)=2I$. Solving, we get,$I _1=3.5I$ and $I _2=1.5I$

The correct option is (b)

Waves from two sources superpose on each other at a particular point amplitude and frequency of both the waves are equal. The ratio of intensities when both waves reach in the same phase and they reach with the phase difference of $90^{\circ}$ will be

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

  1. $1 : 1$

  2. $\sqrt {2} : 1$

  3. $2 : 1$

  4. $4 : 1$

Show answer
Correct Option: C
Explanation:
Relation between intensity $f$
Amplitude
$I= kA^{2} \cos^{2} \left( \dfrac{|theta}{2} \right)$
When both the waves reach in the same phase, then 
$\theta= 0^{o}$ 
$I _{1}= kA^{2} \cos^{2} \left( \dfrac{0^{o}}{2} \right)$
$I _{1} = kA^{2} [\cos 0^{o}=1]$ ------- $(1)$
When both the waves reach with phase difference $90^{o}$.
$\theta = 90^{o}$
$I _{2}= kA^{2} \cos^{2} \left( \dfrac{90^{o}}{2} \right)$
$I _{2} = kA^{2} \cos^{2} (45^{o})$
$I _{2}= kA^{2} \left( \dfrac{1}{2} \right)$ ----- $(2)$
$\dfrac{I _{1}}{I _{2}}= \dfrac{kA^{2}}{kA^{2} (1/2)} =\dfrac{2}{1}$