Question

A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves the end up and down through a distance by 2.0 cm. The motion of the bar is continuous and is repeated regularly 125 times per sec. If the distance between adjacent wave crests is observed to 15.7 cm and the wave is moving along the +ve x-direction, and at \[t = 0\], the element of the string at \[x = 0\] is at mean position \[y = 0\] and is moving downwards, the equation of the wave is best described by: (use\[\pi = 3.14\])
A. \[y = \left( {1\,cm} \right)\sin \left[ {\left( {40.0\,rad/m} \right)x - \left( {785\,rad/s} \right)t} \right]\]
B. \[y = \left( {2\,cm} \right)\sin \left[ {\left( {40.0\,rad/m} \right)x - \left( {785\,rad/s} \right)t} \right]\]
C. \[y = \left( {1\,cm} \right)\cos \left[ {\left( {40.0\,rad/m} \right)x - \left( {785\,rad/s} \right)t} \right]\]
D. \[y = \left( {2\,cm} \right)\cos \left[ {\left( {40.0\,rad/m} \right)x - \left( {785\,rad/s} \right)t} \right]\]

Answer 1

Hint: Refer to the equation of transverse sinusoidal wave. The distance between the adjacent crests is the wavelength of the wave. Use the relation between wavenumber and wavelength to determine the wave number. Substitute these quantities in the equation of the wave.

Complete step by step answer:
We know the equation of transverse sinusoidal wave,
\[y = A\sin \left( {kx - \omega t + \phi } \right)\] …… (1)
Here, A is the amplitude of the wave, \[\omega \] is the angular velocity, t is the time and \[\phi \] is the phase difference.
We know that the angular frequency \[\omega \] is,
\[\omega = 2\pi f\]
Here, \[f\] is the frequency of the wave.
We have given the string repeated regularly 125 times per second. This is the frequency of the wave. The angular frequency of this wave is,
\[\omega = 2\left( {3.14} \right)\left( {125} \right)\]
 The wave number of the transverse wave is given by,
\[k = \dfrac{{2\pi }}{\lambda }\]
We have given the distance between the adjacent crests is 15.7 cm. This is the wavelength of the wave.
Therefore, the wavenumber will become,
\[k = \dfrac{{2\left( {3.14} \right)}}{{0.157\,m}}\]
\[\therefore k = 40\,rad/m\]
In the given question, we have given that the string moves the end up and down through a distance by 2.0 cm. Therefore, the amplitude of the wave is 2.0 cm.
We have given, at \[t = 0\], the element of the string at \[x = 0\] is at the mean position \[y = 0\] and is moving downwards. Therefore, the phase difference \[\phi \] of the wave is zero.
\[\phi = 0\]
Substitute 2 cm for A, \[40\,rad/m\] for k, \[785\,rad/s\] for \[\omega \] and 0 for \[\phi \] in equation (1).
 \[y = \left( {2\,cm} \right)\sin \left( {\left( {40\,rad/m} \right)x - \left( {785\,rad/s} \right)t + 0} \right)\]
\[y = \left( {2\,cm} \right)\sin \left[ {\left( {40\,rad/m} \right)x - \left( {785\,rad/s} \right)t} \right]\]

So, the correct answer is “Option B”.

Note:
since the initial x and y position of the wave is 0, there is no phase difference in the wave. If the x and y position of the wave is different, you can use the relation \[\phi = \dfrac{{2\pi \Delta x}}{\lambda }\].