Tag: wave motion

Questions Related to wave motion

A long string having a cross-sectional area $0.80 mm^2$ mm2and density, $12.5 g/cc$ is subjected to a tension of $64 N$ along the positive x-axis. One end of this string is attached to a vibrator at $x = 0$ moving in transverse direction at a frequency of $20 Hz$. At $t = 0$, the source is at a maximum displacement $y = 1.0 cm.$ What is the velocity of this particle at the instant when $x=50\ cm$  and time $t=0.05\  s$?

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

  1. $y(0.5m,0.05s)=98cm/s$

  2. $y(0.5m,0.05s)=59cm/s$

  3. $y(0.5m,0.05s)=89cm/s$

  4. $y(0.5m,0.05s)=99cm/s$

Show answer
Correct Option: C
Explanation:
Mass per unit length of the string=$\mu=\rho A=0.8\times 10^{-6}\times 12.5\times 10^{3}=0.01kg/m$
Thus speed of the wave=$\sqrt{\dfrac{T}{\mu}}=\sqrt{\dfrac{64}{0.01}}=80m/s$

Amplitude of the wave=A=1cm
$\omega=2\pi\nu=40\pi s^{-1}$
$v=\dfrac{\omega}{k}$
$\implies k=\dfrac{40\pi}{80}=\dfrac{\pi}{2} m^{-1}$
Thus the wave equation is $y=Acos(\omega t-kx)$
$=(1cm)cos[(40\pi s^{-1})t-(\dfrac{\pi}{2}m^{-1})x]$
Hence velocity of a particle=$-\dfrac{dy}{dt}=-\omega A sin(\omega t-kx)$
Thus $y(0.5m,0.05s)=89cm/s$

Transverse waves on a string have wave speed $8.00$ m/s, amplitude $0.0700\  m$ and wavelength $0.32\  m$. The waves travel in the negative x-direction and $t = 0$ the $x = 0$ end of the string has its maximum upward displacement. Write a wave function describing the wave.

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

  1. $\displaystyle \,y\,(x,\,t)\,=\,(0.07\,m)\,sin\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$

  2. $\displaystyle \,y\,(x,\,t)\,=\,(77\,m)\,cos\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$

  3. $\displaystyle \,y\,(x,\,t)\,=\,(0.7\,m)\,sin\,4\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$

  4. $\displaystyle \,y\,(x,\,t)\,=\,(0.97\,m)\,sin\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$

Show answer
Correct Option: A
Explanation:

A left travelling transverse wave given by $y=Asin(kx+\omega t)$

where wave number,$k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{0.32}rad/m$
Speed of wave=$\lambda\nu=8m/s$
$\implies \nu=\dfrac{8}{0.32}Hz=25Hz$
$\implies \omega=2\pi\nu=\dfrac{2\pi}{0.04s}$
Thus $y=(0.07m)sin2\pi(\dfrac{x}{0.32m}+\dfrac{t}{0.04s})$

Transverse waves on a string have wave speed $12.0$ m/s, amplitude $0.05\  m$ and wavelength $0.4\  m$. The waves travel in the $+ x$ direction and at $t = 0$, the $x = 0$ end of the string has zero displacement and is moving upwards. Find the transverse displacement of a point at x = 0.25 m at time t = 0.15 s.

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

  1. $-4.54 \  cm$

  2. $-5.54 \  cm$

  3. $-3.54 \  cm$

  4. $-9.54 \  cm$

Show answer
Correct Option: C
Explanation:

A wave traveling in +x direction is represented by $y=Asin(\omega t-kx)$

where $\omega=2\pi \nu$
and $k=\dfrac{2\pi}{\lambda}$
We know that speed of wave=$v=\lambda\nu$
Thus here
$\omega=2\pi\times \dfrac{v}{\lambda}=2\pi\times \dfrac{12}{0.4}=60\pi s^{-1}$
and $k=\dfrac{2\pi}{0.4}=5\pi m^{-1}$
Thus wave is $y=(0.05m)sin((60\pi s^{-1})t-(5\pi m^{-1})x)$
Thus the displacement of point at $x=0.25m$ and $t=0.15s$ can be found by putting the values in the equation of wave.
Thus $y(x=0.25m,t=0.15s)=-0.0354m=-3.54cm$

A long string having a cross-sectional area $0.80 mm^2$ mm2and density, $12.5 g/cc$ is subjected to a tension of $64 N$ along the positive x-axis. One end of this string is attached to a vibrator at $x = 0$ moving in transverse direction at a frequency of $20 Hz$. At $t = 0$, the source is at a maximum displacement $y = 1.0 cm.$ What is the displacement of the particle of the string at $x = 50 cm$ at time $t = 0.05 s$ ?

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

  1. $0.71 cm $

  2. $0.91 cm $

  3. $0.58 cm $

  4. $0.31 cm $

Show answer
Correct Option: A
Explanation:
Mass per unit length of the string=$\mu=\rho A=0.8\times 10^{-6}\times 12.5\times 10^{3}=0.01kg/m$
Thus speed of the wave=$\sqrt{\dfrac{T}{\mu}}=\sqrt{\dfrac{64}{0.01}}=80m/s$

Amplitude of the wave=A=1cm
$\omega=2\pi\nu=40\pi s^{-1}$
$v=\dfrac{\omega}{k}$
$\implies k=\dfrac{40\pi}{80}=\dfrac{\pi}{2} m^{-1}$
Thus the wave equation is $y=Acos(\omega t-kx)$
$=(1cm)cos[(40\pi s^{-1})t-(\dfrac{\pi}{2}m^{-1})x]$
Thus $y(0.5m, 0.05s)=0.71cm$

Three component sinusoidal waves progressing in the same direction along the same path have the same period, but their amplitudes are $A$, $\displaystyle \frac{A}{2}$ and $\displaystyle \frac{A}{3}$ respectively. The phase of the variation at any position $x$ on their path at time $t = 0$ are $0$, $\displaystyle -\frac{\pi}{2}$ and $-\pi$ respectively. Find the amplitude and phase of the resultant wave.

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

  1. $\displaystyle \frac{5}{6} A$, $\displaystyle -tan^{-1} \left (\frac{3}{4} \right )$

  2. $\displaystyle \frac{7}{6} A$, $\displaystyle -tan^{-1} \left (\frac{3}{4} \right )$

  3. $\displaystyle \frac{5}{6} A$, $\displaystyle -tan^{-1} \left (\frac{1}{4} \right )$

  4. $\displaystyle \frac{7}{6} A$, $\displaystyle -tan^{-1} \left (\frac{1}{4} \right )$

Show answer
Correct Option: A
Explanation:

The waves with opposite phases superimpose to give a resultant wave of amplitude, $A-\dfrac{A}{3}=\dfrac{2A}{3}$ with phase angle $0$ at time $t=0$.

This superimposes with wave of amplitude $\dfrac{A}{2}$ in with phase $-\dfrac{\pi}{2}$ at $t=0$.

Hence, the resulting wave has amplitude $\sqrt{(\dfrac{2A}{3})^2+(\dfrac{A}{2})^2}=\dfrac{5}{6}A$
The phase of the resulting wave is $tan^{-1}\dfrac{-\dfrac{A}{2}}{\dfrac{2A}{3}}$$=-tan^{-1}\dfrac{3}{4}$

Two wave pulses travel in opposite directions on  a string and approach each other. The shape of one pulse is inverted with respect to the other.

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

  1. The pulse will collide with each other and vanish

    after collision.

  2. The pulses will reflect each other, that is pulse

    going towards right will finally move towards left

    and vice versa.

  3. The pulses will pass through each other but their

    shapes will be modified.

  4. The pulses will pass through each other without

    any change.

Show answer
Correct Option: D

If 'v' is the velocity of sound in a gas then 'v' is directly proportional to (where M, d and T represents molecular weight of gas, density of gas and its temperature respectively.)

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

  1. $\sqrt{M}$

  2. $\displaystyle \frac{1}{\sqrt{d}}$

  3. $\sqrt{T}$

  4. Both (2) and (3)

Show answer
Correct Option: D

 Standing waves are generated on string laded with a cylindrical body. If the cylinder immersed in water, the length of the loops changes by a factor of 2.2. The specific gravity of the material of the cylinder is 

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

  1. 1.11

  2. 2.15

  3. 2.50

  4. 1.26

Show answer
Correct Option: A

In a string the speed of wave is 10 m/s and its frequency is 100 Hz . The value of the phase difference at a distance 2.5 cm will be :

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

  1. ${ \pi }/{ 2 }$

  2. ${ \pi }/{ 8 }$

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

  4. ${ 2\pi }$

Show answer
Correct Option: A
Explanation:

Speed of wave $(v) = 10 m/s$

Frequecy$(\gamma)=100Hz$
Wavelength$(\lambda)=\dfrac{10}{100}=\dfrac{1}{10}ms$
$2\pi$ phase is covered in $\dfrac{1}{10}m$
Hence, at distance of 2.5 m, the phase is $\dfrac{2\pi \times 0.025}{0.1}=\dfrac{\pi}{2}$



A string of mass $3$kg is under tension of $400$N. The length of the stretched string is $25$cm. If the transverse jerk is stuck at one end of the string find the velocity?

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

  1. $25 \pi^2 m s^{-2}$

  2. $-5 \pi^2 m s^{-2}$

  3. $5 \pi^2 m s^{-2}$

  4. $-25 \pi^2 m s^{-2}$

Show answer
Correct Option: B
Explanation:

Here, A = $5cm = 0.05$m, T = $0.2$s


$\therefore \omega = \dfrac{2 \pi}{T} = \dfrac{2 \pi}{0.2} = 10 \pi rad s^{-1}$

Velocity and acceleration of the particle executing SHM at any displacement xi given by

Velocity, $v = \omega \sqrt{A^2 - x^2}$
and acceleration, $a = - \omega^2 x$
when $x = 5 cm = 0.05$m
$\therefore v = 10 \pi \sqrt{(0.05)^2 - (0.05)^2} = 0 and a = - (10 \pi)^2(0.05) = -5 \pi^2 m s^{-2}$