Tag: wave velocity

Questions Related to wave velocity

A wave propagates on a string in positive $x-$ direction with a speed of $40\ cm/s$. The shape of string at $t=2\ s$ is $y=10\cos \,\dfrac{x}{5}$, where $x$ and $y$ are in centimetre. The wave equation is :

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

  1. $y=10\cos \left(\dfrac{x}{5}-8t\right)$

  2. $y=10\sin \left(\dfrac{x}{5}-8t\right)$

  3. $y=10\cos \left(\dfrac{x}{5}-8t+16\right)$

  4. $y=10\sin \left(\dfrac{x}{5}-8t+16\right)$

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Correct Option: C

A wave pulse is propagating with speed $c$ towards positive $x-$axis. The shape of pulse at $t=0$, is $y=ae^{-x/b}$ where $a$ and $b$ are constant. The equation of wave is :

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

  1. $ae^{-\left(\dfrac{x-ct}{b}\right)}$

  2. $ae^{\dfrac{ct+x}{b}}$

  3. $ae^{x-ct}$

  4. $none\ of\ these$

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Correct Option: A

1 meter long stretched wire of a sonometer vibrates with its fundamental frequency of 256 Hz. If the length of the wire is decreased to 25 cm and the tension remains the same, then the fundamental frequency of vibration will be:-

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

  1. 64 Hz

  2. 256 Hz

  3. 512 Hz

  4. 1024 Hz

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Correct Option: A

In a stretched string, 

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

  1. Only transverse waves can exist

  2. Only longitudinal waves can exist

  3. Both transverse and longitudinal waves can exist

  4. None of these

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Correct Option: A

A travelling wave is propagating along negative $x-$axis through a stretched string. The displacement of a particle of the string at $x=0$ is $y=a\cos \omega t$. The speed of wave is $c$. The wave equation is :

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

  1. $y=a\cos \omega t$

  2. $y=2a\cos \omega t$

  3. $y=a\cos \omega$ $\left(t-\dfrac{x}{c}\right)$

  4. $y=a\cos \left(\omega t+\dfrac{\omega x}{c}\right)$

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Correct Option: D

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

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

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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. $-9.54 \  cm$

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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 $

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

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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}$