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Distinction between interference and beats - class-XI

Description: distinction between interference and beats
Number of Questions: 16
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Tags: waves superposition of waves-1: interference and beats oscillation and waves wave motion physics
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Two sound sources are moving in opposite direction with velocity $v _1$ and $v _2$ $(v _1>v _2)$. Both are moving away from a stationary observer.the frequency of both the source is $900\ Hz$. What is the value of $v _1 - v _2 $  so that the beat frequency observed will be $6\ Hz$  ?


Speed of sound =$300\ ms^{-1}$

  1. $1\ ms^{-1}$

  2. $4\ ms^{-1}$

  3. $3\ ms^{-1}$

  4. $2\ ms^{-1}$


Correct Option: D
Explanation:

$f _1 = 900(\dfrac{300}{300+v _1})$

$\implies f _1= 900(1+ \dfrac{v _1}{300})^{-1}$
$\implies f _1 = 900 - 3v _1$
Similarly 
$f _2 = 900 - 3v _2$
So,
$f _1 -f _2 = 6$
$3(v _1 - v _2)= 6$
$\implies v _1 - v _2 = 2\ ms^{-1}$

Sound waves of equal amplitude wave frequency $(n-1),n _1(nH^1)$. They superinpose to give Beat. The no of Beats poduced per sec cuPll be:

  1. l

  2. 0

  3. 3

  4. 2


Correct Option: D

A sources of sonic oscillations with frequency n= $1700$ Hz and a receiver are located on the same normal to a wall. Both the source and receiver are stationary, and the wall recedes from the source with velocity u= $6.0$ cm/s. Find the beat frequency registred by the receiver. The velocity of sound is equal to $v= 340$ m/s.

  1. $0.2$ Hz

  2. $0.3$ Hz

  3. $0.4$ Hz

  4. $0.6$ Hz


Correct Option: D

Two sound sources (of same frequency ) are placed at distance of 100 meter. An observer, when moving between both sources, hears 44 beats per second. The distance between sound source is now changed to 400 meter then the beats/second heard by observer will be  :

  1. 2

  2. 4

  3. 8

  4. 16


Correct Option: D

Two open organ pipes 80 and 81 cm long found to give 26 beats in 10 sec, when each is sounding its fundamental note. Then the velocity of sound in air is

  1. 337 $m s ^ { - 1 }$

  2. 370 $m s ^ { - 1 }$

  3. 345 $m s ^ { - 1 }$

  4. 350 $m s ^ { - 1 }$


Correct Option: A

Two monochromatic light waves of amplitudes $A$ and $2A$ interfering at a point, have a phase difference of ${60^0}.$ The intensity at that point will be  proportional to :

  1. $3{A^2}$

  2. $5{A^2}$

  3. $7{A^2}$

  4. $9{A^2}$


Correct Option: C

Two sound waves with wavelength $5$m and $5.5$ m respectively. each propoggate in a gas with velocity $300$ m/s. we expect the following number of beats per second

  1. 12

  2. 0

  3. 1

  4. 6


Correct Option: D

Two waves are approaching each other with a velocity of $16\, m/s$ and frequency $n$. the distance between two consecutive nodes is 

  1. $\dfrac{16}{n}$

  2. $\dfrac{8}{n}$

  3. $\dfrac{n}{16}$

  4. $\dfrac{n}{8}$


Correct Option: A

A body is walking away from a wall towards an observe at a speed of 1 m/s and blows a whistle whose frequency is 680 Hz. The number of beats heard by the observe per second is approximately.(velocity of sound in air = 340 m/s)

  1. 4

  2. 8

  3. 2

  4. zero


Correct Option: A

Two sound waves of equal intensity I produce beats . The maximum intensity of sound produced in beats will be

  1. I

  2. 4I

  3. 2I

  4. I/2


Correct Option: C

$y _1 = A cos (2f _1t)$     and $y _2 = A cos (2 f _2t),$, then $y _{total} $ is

  1. $y _{total} = y _1 + y _2 = A {cos (2 f _1t) + cos (2 f _2t)}$

  2. $y _{total} = y _1 - y _2 = A {cos (2 f _1t) - cos (2 f _2t)}$

  3. $y _{total} =\dfrac{ y _1}{ y _2} = A\dfrac{{cos (2 f _1t)}}{{cos (2 f _2t)}}$

  4. $y _{total} = y _1 \times y _2 = A {cos (2 f _1t) \times cos (2 f _2t)}$


Correct Option: A
Explanation:

Resultant  $(y _{total})$ of the two waves is equal to the superposition of the waves and is given by,
$y _{total}  = y _1+y _2$
$\therefore$  $y _{total} = A \ cos(2f _1t)+ A \ cos(2f _2t)$

A tuning fork of frequency 240 Hz produce 5 beats with a sonometer wire. On increasing the tension in the wire if the number of beats changes to 4 per second, the initial frequency of the wire is

  1. 8285 Hz

  2. 275 Hz

  3. 376 Hz

  4. 264 Hz


Correct Option: A

Two waves of wavelengths 99 cm and 100 cm both travelling with velocity 396 m/s are made of interfere. The number of beats produced by them per second are

  1. $1$

  2. $2$

  3. $4$

  4. $8$


Correct Option: C
Explanation:

$\begin{array}{l} Velocity\, \, of\, \, wave\, \, V=n\lambda  \ Where\, \, n=frequency\, \, of\, \, wave\, \,  \ \Rightarrow n=\frac { v }{ \lambda  }  \ { n _{ 2 } }=\frac { { { v _{ 2 } } } }{ { { \lambda _{ 2 } } } } =\frac { { 396 } }{ { 100\times { { 10 }^{ -2 } } } } =396Hz \ no.\, \, of\, \, beats\, \, ={ n _{ 1 } }-n _2\, =4 \end{array}$

If 2 waves of same frequency and same amplitude on superposition, produce a resultant disturbance of the same amplitude, the waves differ in phase by

  1. $ \dfrac { \pi }{ 3 } $

  2. $ \dfrac { 2\pi }{ 3 } $

  3. $ { \pi } $

  4. $ { 3\pi } $


Correct Option: B
Explanation:

Let the amplitude of each wave be $A$ i.e.  $A _1 = A _2 = A$
Thus amplitude of resultant wave  $A _R =A$
Using   $A^2 _R =  A _1^2 + A _2^2 + 2A _1A _2 \cos\theta$
$\therefore$  $A^2 =  A^2 + A^2 + 2A^2 \cos\theta$
Or  $\cos\theta = -0.5$
$\implies  \ \theta = \dfrac{2\pi}{3}$

Two identical wires are stretched by the same tension of $101$ N & each emits a note of frequency $202$ Hz. If the tension in one wire is increased by $1N$ , then the beat frequency is:

  1. $2 Hz$

  2. $\frac{1}{2}Hz$

  3. $1 Hz$

  4. None of these


Correct Option: C
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