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Standing waves - class-XI

Description: standing waves
Number of Questions: 16
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Tags: physics waves
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Which of the following function represent traveling waves?

  1. $ y = (x + 5t)^3 $

  2. $ y = tan (2x + 3t) $

  3. $ y = \theta^{(4t+2x)^2} $

  4. $ y = \frac { 1 }{ x + 3t } $


Correct Option: D

A progressive wave is incident normally on a flat reflector. The reflected wave overlaps with the incident wave and a stationary wave is formed.
At an antinode, what could be the ratio $\dfrac{displacement of the incident wave}{displacement of the reflected wave}$ at any instant?

  1. $-1$

  2. $0$

  3. $1$

  4. $2$


Correct Option: C
Explanation:

Standing wave is formed when two waves with equal magnitude and and opposite direction overlap Both waves have same amplitude at antinode and opposite amplitude at node.

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 X = $\dfrac { v\lambda }{ 2 } $;n = 0,1,2,..................

  2. A standing wave have nodes at X=$\left( n+\dfrac { 1 }{ 2 } \right) \dfrac { \lambda }{ 2 } $;n = 0,1,2,...............................

  3. A wave travelling along +x direction.

  4. A wave travelling along -x direction


Correct Option: B
Explanation:

Intensity wave is standing wave and have nodes at 
$X=\left( n+\dfrac { 1 }{ 2 }  \right) \dfrac { \lambda  }{ 2 } $ n=0,1,2,......

A sine wave described by the equation $x = 2 sin (2 \pi t-3 x)$ is progressing along an x axis. In order that a standing wave is setup, what should be the equation of the reflected wave

  1. $x = 2 sin (2 \pi t-3 x)$

  2. $x = 2 sin (2 \pi t+3 x)$

  3. $x = 2 sin (2 \pi t-3 x+ \pi)$

  4. $x = 2 sin (2 \pi t-3 x+\pi/3)$


Correct Option: C
Explanation:

Since the reflected wave differs from the incident wave by two aspects (direction of the wave and the phase of the wave), the wave will have an equation given by $x = 2 sin (2 \pi t-3 x +\pi)$ to form a stationary wave

The correct option is (c)

A string attached to a tuning fork of frequency 300 Hz is made to vibrate. The other end of the string is fixed to a wall. If stationary waves are to be set up, what should be the phase of the reflected wave

  1. $\pi $ rads

  2. $\pi/2 $ rads

  3. $\pi/3 $ rads

  4. none of the above


Correct Option: A
Explanation:

The reflected and the actual wave has to be out of phase and only in such cases, a stationary wave is being setup

The correct option is (a)

Two sine waves of same frequency (f) and amplitude (A) are superimposed from opposite directions along a straight line. The resultant wave will have an amplitude of 

  1. A

  2. A/2

  3. 2A

  4. 3A/2


Correct Option: C
Explanation:

Their amplitudes get doubled

The correct option is (c)

A traveling wave passes a point of observation. At this point, the time interval between successive crests is 0.2 seconds and  

  1. The wavelength is 5 m

  2. The frequency is 5 Hz

  3. The velocity of propagation is 5 m/s

  4. The wavelength is 0.2 m


Correct Option: B
Explanation:

The frequency is $5Hz$


Phase difference,

$\Delta \phi =\dfrac{2\pi}{T}(\Delta t)$

$2\pi =\dfrac{2\pi}{T}(0.2)$

$\Rightarrow \dfrac{1}{T}=5 sec^{-1}$

$n=5Hz$

A string is vibrating in $n$ loops. The number of nodes and antinodes respectively are

  1. $n, n$

  2. $(n+1), n$

  3. $n, (n-1)$

  4. $(n-1), n$


Correct Option: B

An organ pipe of length $80\ cm$ is opened at $x=0$ and closed at $x=80\ cm$. Speed of sound in the air column is $320\ m/sec$. If standing waves are generated in the closed organ pipe, then the correct equation of standing waves is/are (Here $s=$ longitudinal displacement, $P _{ex}=$ pressure excess) (Neglect the end correction).

  1. $S=A\cos\left(\dfrac{5\pi}{4}x\right)\sin\left(400\pi t\right)$

  2. $S=A\cos\left(\dfrac{5\pi}{8}x\right)\cos\left(1000\pi t\right)$

  3. $P _{ex}=A\cos\left(\dfrac{5\pi}{8}x\right)\sin\left(200\pi t\right)$

  4. $P _{ex}=A\sin\left(\dfrac{25\pi}{8}x\right)\cos\left(1000\pi t\right)$


Correct Option: C

Consider single slit experiment of diffraction of light. If light of wavelength $5000\ \mathring { A } $ fall on a slit of width $1\mu\ m$ then the angular width of central maximum. 

  1. ${30}^{o}$

  2. ${15}^{o}$

  3. ${60}^{o}$

  4. $None\ of\ these$


Correct Option: A

The equation of a traveling and stationary wave are ${ y } _{ 1 }=a sin(\omega t-kx)$ and ${ y } _{ 2 }=a \sin kx  \cos \omega t$. The phase difference between two point ${ x } _{ 1 }=\dfrac { \pi  }{ 4k }$ and $ { x } _{ 2 }=\dfrac { 4\pi  }{ 3k } $ are ${ \phi  } _{ 1 }$ and ${ \phi  } _{ 2 }$ respectively for two waves where k is the wave number, the ratio of ${ \phi  } _{ 1 }/{ \phi  } _{ 2 }$ 

  1. 6/7

  2. 16/3

  3. 12/13

  4. 13/12


Correct Option: D

A standing wave pattern is formed on a string. One of the waves is given by equation  $Y _ { 1 } a \cos ( \omega t - K X + \pi / 3 )$  then the equation of the other wave such at  $X = 0$  a noode is formal

  1. $y _{ 2 } = \operatorname { a sin } \left( \omega t + K X + \dfrac { \pi } { 3 } \right)$

  2. $y _ { 2 } = a \cos \left( \omega t + K X + \dfrac { \pi } { 3 } \right)$

  3. $y _ { 2 } = a \cos \left( \omega t + K X + \dfrac { 2 \pi } { 3 } \right)$

  4. $y _ { 2 } = a \cos \left( \omega t + K X + \dfrac { 4 \pi } { 3 } \right)$


Correct Option: D

Two simple harmonic waves of amplitude 5 cm and 3 cm and of the same frequency travelling with the same speed in opposite directions superpose to produce stationary waves. The ration of the amplitude at a node to that at an antinode in the resultant wave is

  1. zero

  2. infinity

  3. 5:3

  4. 1:4


Correct Option: A

The equation of stationary wave is given by $y=5\, cos (\pi x/3)\, sin 40 \pi t$ where y and x are given in cm and time t in second. Then a node occurs at the following distance 

  1. 3 cm

  2. 10 cm

  3. 5 cm

  4. 1.5 cm


Correct Option: A

A $string$ is stretched between fixed points separated by $75.0\ cm$. It is observed to have resonant frequencies of $420\ Hz$ and $315\ Hz$. There are no other resonant frequencies between these two.
Then, the lowest resonant frequency for this string is :

  1. $1.05$Hz

  2. $1050$Hz

  3. $10.5$Hz

  4. $105$Hz


Correct Option: D
Explanation:

Given,  $\dfrac { nV }{ 2L } =315\quad \longrightarrow (1)$

     &     $\dfrac { \left( n+1 \right) V }{ 2L } =420\quad \longrightarrow (2)$
equation (2) $-$ equation (1), we get
$\dfrac { \left( n+1 \right) V }{ 2L } -\dfrac { nV }{ 2L } =420-315$
$\Rightarrow \quad \left[ \dfrac { V }{ 2L } =105\quad { H } _{ 3 } \right] \rightarrow $  Lowest possible resonant frequency

$\therefore $  Option (D) is correct.

A wave represented by $y=2 cos (4x-\pi t)$ is superposed with another wave to form a stationary wave such that the point x= 0 is a node. The equation of other wave is:

  1. $2 sin(4x+\pi t)$

  2. $-2 cos (4x -\pi t)$

  3. $-2 cos (4x +\pi t)$

  4. $-2 sin (4x -\pi t)$


Correct Option: C
Explanation:

According to options

If $y _2=-2\cos(4x+\pi t)$
Then, when superimposed,
$y=y _1+y _2\ \quad 2\cos(4x-\pi t)-2\cos(4x+\pi t)\ =2[2\sin(\cfrac{(4x-\pi t)+(4x+\pi t)}{2})\sin(\cfrac{(4x-\pi t)-(4x+\pi t)}{2})]\ \quad=2[2\sin(4x)\sin(-\pi t)]\y=-4\sin(4x)\sin(\pi t)$
at $y=0\Rightarrow y=0$ (i.e node)

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