Tag: wave velocity

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Questions Related to wave velocity

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

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:-

  1. 64 Hz

  2. 256 Hz

  3. 512 Hz

  4. 1024 Hz
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The fundamental frequency of a stretched wire is inversely proportional to its length, f proportional to 1 / L, when tension and mass density remain constant. When the length is decreased from 1 m (100 cm) to 25 cm, the length is reduced by a factor of 4, so the fundamental frequency increases by a factor of 4. Thus, f2 = 256 * 4 = 1024 Hz.

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. $y(0.5m,0.05s)=98cm/s$
  2. $y(0.5m,0.05s)=59cm/s$
  3. $y(0.5m,0.05s)=89cm/s$
  4. $y(0.5m,0.05s)=99cm/s$
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. $\displaystyle \,y\,(x,\,t)\,=\,(0.07\,m)\,sin\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$
  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. $\displaystyle \,y\,(x,\,t)\,=\,(0.97\,m)\,sin\,2\,\pi\,\left ( \frac{x}{0.32\,m}\,+\,\frac{t}{0.04\,s} \right )$
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. $-4.54 \ cm$
  2. $-5.54 \ cm$
  3. $-3.54 \ cm$
  4. $-9.54 \ cm$
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. $0.91 cm $
  3. $0.58 cm $
  4. $0.31 cm $
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. $\displaystyle \frac{5}{6} A$, $\displaystyle -tan^{-1} \left (\frac{3}{4} \right )$
  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. $\displaystyle \frac{7}{6} A$, $\displaystyle -tan^{-1} \left (\frac{1}{4} \right )$
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}$