Tag: wave motion

Questions Related to wave motion

The $(x, y)$ co-ordinates of the corners of a square plate are $(0, 0) (L, 0) (L, L)$ & $(0, L)$. The edges of the plate are clamped & transverse standing waves are set up in it. If $u (x, y)$ denotes the displacement of the plate at the point $(x, y)$ at some instant of time, the possible expression(s) for $u$ is/are : ($a$ = positive constant) 

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $a\displaystyle \cos \left(\dfrac{\pi x}{2 L}\right)$ $\displaystyle \cos \left(\dfrac{\pi y}{2 L}\right)$

  2.  $a\displaystyle \sin \left(\dfrac{\pi x}{L}\right)$ $\displaystyle \sin \left(\dfrac{\pi y}{L}\right)$

  3. $a\displaystyle \sin \left(\dfrac{\pi x}{L}\right)$ $\displaystyle \sin \left(\dfrac{2\pi y}{L}\right)$

  4.  $a\displaystyle \cos \left(\dfrac{2\pi x}{L}\right)$ $\displaystyle \sin \left(\dfrac{\pi y}{L}\right)$

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Correct Option: B,C
Explanation:
The expression for $u(x,y)$ should satisfy the following conditions-
i) $u=0$ at $x=0$ and at $y=0$
ii) $u=0$ at $x=L$ and at $y=L$
Only choices B and C satisfy this condition.

The displacement of the particle at $x=0$ of a stretched string carrying wave in the positive x-direction is given $f(t)=A sin \frac {t} {T})$. The wave speed is V. Write the wave equation 

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $f(x,t)=A sin (\frac {t} {T}) - (\frac{x} {V})$

  2. $f(x,t)=A sin (\frac {t} {T}) + (\frac{x} {VT})$

  3. $f(x,t)=A sin (t+- (\frac{x} {V})$

  4. $f(x,t)=A sin (\frac {t} {T}) - (\frac{x} {VT})$

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Correct Option: D

A uniform string of length $L$ fixed between the two ends is vibrating in three segments. The wavelength of wave in string is

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

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

  2. $3L$

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

  4. $\dfrac { 3L }{ 2 } $

Show answer
Correct Option: C
Explanation:

$\begin{array}{l} \dfrac { { 3\lambda  } }{ 2 } =L \ \lambda =\dfrac { { 2l } }{ 3 }  \end{array}$

$\therefore $ Option $C$ is correct.

A uniform rope of length $L$ and mass ${m _1}$ hangs vertically from a rigid support. A block of mass ${m _{2\,}}$ is attached to the free end of the rope. A transverse pulse of wavelength ${\lambda _1}$ is produced at the lower end of the rope. The Wavelength of the pulse when it reaches the top of the rope is ${\lambda _2}$. The ratio ${\lambda _2}/{\lambda _1}$ is 

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $\sqrt {\frac{{{m _1} + {m _2}}}{{{m _1}}}} $

  2. $\sqrt {\frac{{{m _1}}}{{{m _2}}}} $

  3. $\sqrt {\frac{{{m _1} + {m _2}}}{{{m _2}}}} $

  4. $\sqrt {\frac{{{m _2}}}{{{m _1}}}} $

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Correct Option: C

A stretched string of length $1m$ fixed at both ends, having a mass of $5\times{10}^{-4}kg$ is under a tension of $20N$. It is plucked at a point situated at $200cm$ from one end. The stretched string would vibrate with a frequency of

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $200Hz$

  2. $100Hz$

  3. $250Hz$

  4. $256Hz$

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Correct Option: A
Explanation:

$\begin{array}{l} v=\dfrac { 2 }{ { 2l } } \sqrt { \dfrac { { T\times l } }{ M }  }  \ or\, \, ,\, \, \sqrt { \dfrac { T }{ { Ml } }  } =\sqrt { \dfrac { { 20 } }{ { 5\times 10-4\times l } }  }  \ =\sqrt { 4\times { { 10 }^{ 4 } } } Hz=200Hz \end{array}$

 A solid cylinder of mg 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to and hanging freely. Tension in the string required to produce angular acceleration of revolutions $s ^ { - 2 }$ is

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. 78.5 N

  2. 157 N

  3. 25 N

  4. 50 N

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Correct Option: B
Explanation:

Mass=m=50kg

Radius=0.5m
Angular acceleration$=\alpha=2 rev/s^{2}$
Torque$=T \times R = T \times 0.5= \cfrac{T}{2} Nm ------(i)$
We know, $T=1 \alpha------(ii)$
From (i) and (ii),
$\cfrac{T}{2}=1 \alpha = \left( \cfrac{MR^{2}}{2} \right) \times (2 \times 2 \pi) rad/s^{2}$
$\therefore 1$ solid cylinder$=\cfrac{MR^{2}}{2}$
$\cfrac { T }{ 2 } =\cfrac { 50\times { (0.5) }^{ 2 } }{ 2 } \times 4\pi =50\pi =157N$

The vibration of a string of length 60 cm fixed at both ends are represented by $ y=4sin (\frac { \pi x}{15}) cos (96 \pi t) $ where x and y are in cm and t in second. the particle velocity at x=7.5 cm and t=0.25 s is

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. Zero

  2. $ 10 cm s^{-1} $

  3. $ 100 cm s^{-1} $

  4. $ (4 \times 96) cm s^{-1} $

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Correct Option: A

A $100$ Hz sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of $3.5 \times 10^{-3}$ kg/m and a tension of $35$ N. At time t = 0, the point x = 0 has zero displacements and the slope of the string is $\pi/20$. Then select the wrong alternative

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. Velocity of wave is $100$ m/s

  2. Angular frequency is $(200 \pi)$ rad /s

  3. Amplitude of wave is $0.025$ m

  4. Propagation constant is $(4 \pi)$ $m^{-1}$

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Correct Option: D

A uniform string fixed at both ends is vibrating in 3rd harmonic and equation $y = 4 ( \mathrm { cm } )$ $\sin \left[ \left( 0.8 \mathrm { cm } ^ { - 1 } \right) \times \right] \cos \left[ \left( 400 \pi \mathrm { s } ^ { - 1 } \right) t \right]$The length of the vibrating string is

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. $6.75 \mathrm { m }$

  2. $12.45 \mathrm { m }$

  3. $11.8 \mathrm { m }$

  4. $18.7 \mathrm { m }$

Show answer
Correct Option: C
Explanation:

$ y=4\sin  \left( { 0.8x } \right) \cos  \left( { 400\pi t } \right)  \ w=400\pi =2\pi f \ f=200\, Hz \ v=\dfrac { w }{ k } =\dfrac { { 400\pi \times 100 } }{ { 0.8 } } m/s \ 3\cdot \dfrac { v }{ { 2l } } =f \ \Rightarrow 200=\dfrac { { 3\times 400\pi \times 100 } }{ { 2\times l\times 0.8 } }  \ \Rightarrow l=\dfrac { { 3\times 400\pi  } }{ { 4\times 0.8 } }  \ =\dfrac { { 300\pi  } }{ { 0.8 } } \, cm \ =11.8\, m$

Hence,
option $(C)$ is correct answer.

The wave function for the wave pulse is $ Y (X,t) = \frac {0.1a^3}{a^2 +(X-Vt)^2}  with a = 4 cm. At X = 0 $ The displacement y (x,t) is observed to decreases from its maximum value to half of that value in time $ t = 2 \times 10^{-3} s $ choose the correct statement 

physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

  1. The wave pulse is moving is negative X direction with speed 10 m/s

  2. The wave pulse is moving is positive X direction with speed 10 m/s

  3. The wave pulse is moving is negative X direction with speed 20 m/s

  4. The wave pulse is moving is positive X direction with speed 20 m/s

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Correct Option: A