Tag: speed and acceleration of travelling wave

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Questions Related to speed and acceleration of travelling wave

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

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 :

  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. $y=10\sin \left(\dfrac{x}{5}-8t+16\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Wave speed v = 40 cm/s. At t = 2 s, y = 10 cos(x / 5). The general form is y = 10 cos(x / 5 - omega * t + phi). Since v = omega / k, omega = v * k = 40 * (1 / 5) = 8 rad/s. At t = 2, y = 10 cos(x / 5 - 16 + phi) = 10 cos(x / 5). Thus, phi = 16.

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

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 :

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

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

  3. $ae^{x-ct}$

  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A wave pulse moving in the positive x-direction with speed c has the form f(x - ct). Substituting (x - ct) into the initial shape f(x) = a * e^(-x/b) gives a * e^(-(x - ct) / b).

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

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 :

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

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

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

  4. $y=a\cos \left(\omega t+\dfrac{\omega x}{c}\right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a wave traveling in the negative x-direction, the argument is (omega * t + k * x). Given k = omega / c, the equation is y = a * cos(omega * t + omega * x / c).

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

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 $t=0.05\  s$?

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

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

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

  4. <span>$y(0.5m,0.05s)=99cm/s$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $y(0.5m,0.05s)=89cm/s$

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

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.

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

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

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

  4. <span>$\displaystyle \,y\,(x,\,t)\,=\,(0.97\,m)\,sin\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $y=(0.07m)sin2\pi(\dfrac{x}{0.32m}+\dfrac{t}{0.04s})$

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

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.

  1. $-4.54 \ &nbsp;cm$

  2. <span>$-5.54 \ &nbsp;cm$</span>

  3. <span>$-3.54 \ &nbsp;cm$</span>

  4. <span>$-9.54 \ &nbsp;cm$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $y(x=0.25m,t=0.15s)=-0.0354m=-3.54cm$

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

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 $t = 0.05 s$ ?

  1. $0.71 cm $

  2. <span>$0.91 cm $</span>

  3. <span>$0.58 cm $</span>

  4. <span>$0.31 cm $</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Mass per unit length of the string=$\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 $y(0.5m, 0.05s)=0.71cm$

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

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.

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

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

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

  4. <span>$\displaystyle \frac{7}{6} A$, $\displaystyle -tan^{-1} \left (\frac{1}{4} \right )$</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
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}$