Question

The transverse displacement of a string (clamped at its both ends) is given by \[y(x,t) = 0.06\sin (\dfrac{{2\pi }}{3})x\cos (120\pi t)\] where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10¯²kg. Answer the following : (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave? (c) Determine the tension in the string.

Answer 1

Hint: Standing wave is also known as stationary wave. If a combination of two waves moving in opposite directions, then each having the same amplitude and frequency.

Formula used
Travelling wave equation in positive direction of x-axis:
\[y(x,t) = A\sin (\omega t \pm kx)\]
Here, A is amplitude, \[\omega\] is angular frequency of the wave and k is propagation constant.
Standing wave equation:
\[y(x,t) = A\sin (kx)\cos (\omega t)\]
Speed of wave in a string:
\[v = \sqrt {\dfrac{T}{\mu }} \]
Where \[\mu\] is mass per unit length and T is tension in the string.

Complete step by step solution:
Given Transverse displacement of a string as:
\[y(x,t) = 0.06\sin (\dfrac{{2\pi }}{3})x\cos (120\pi t)\]
Length of the string=1.5 m
Mass=\[3.0 \times {10^{ - 2}}{\rm{ kg}}\]

(a) As we know travelling wave equation is \[y(x,t) = A\sin (\omega t \pm kx + \Phi )\]
Standing wave equation is \[y(x,t) = A\sin (kx)\cos (\omega t)\]
Now compare these equations with the given transverse displacement equation. As the displacement equation is similar to the second equation which is the standing wave. Thus the given wave equation is a stationary wave.

(b) As the stationary wave that is formed by the superposition of two waves moving in opposite directions. So the wavelength, frequency and speed of each wave will be the same.
Let the equation of wave as:
\[{y_1} = r\sin (\omega t - kx)\]
Equation of opposite wave as:
\[{y_1} = r\sin (\omega t + kx)\]
If two waves are travelling in opposite directions then by superposition principle,
\[y = {y_1} + {y_2}\]
Substituting the values of \[{y_1}{\rm{ and }}{{\rm{y}}_2}\] ,
\[y = 2r\sin kx\cos \omega t\] (using the relation \[2\sin A\cos B = \sin (A + B) + \sin (A - B)\] )
Now compare this with the given transverse displacement equation:
\[y(x,t) = 0.06\sin (\dfrac{{2\pi }}{3})x\cos (120\pi t)\]
After comparing we get
Wavelength \[\lambda \] ,
As \[k = \dfrac{{2\pi }}{\lambda } = \dfrac{{2\pi }}{3}\]
\[\lambda = 3m\]
Speed v,
As \[v = \dfrac{\omega }{k}\]
\[vk = \omega \]
\[\Rightarrow v \times \dfrac{{2\pi }}{3} = 120\pi \]
\[\Rightarrow v = 180\,m/s\]
Frequency f,
As \[v = \lambda f\]
\[\begin{array}{l}180 = 3 \times f\\ \Rightarrow f = 60Hz\end{array}\]

(c) To find tension in the string :
As velocity of transverse wave is \[v = \sqrt {\dfrac{T}{\mu }} \]
On squaring on both sides,
\[{v^2} = \dfrac{T}{m}\]
\[\Rightarrow T = {v^2}m\]
As, \[\mu = \dfrac{\text{mass}}{\text{length}}\]
\[\text{mass},m = \dfrac{{3 \times {{10}^{ - 2}}}}{{1.5}} = 2 \times {10^{ - 2}}\]
\[\begin{array}{l}{\rm{ }} = {(180)^2} \times (2 \times {10^{ - 2}})\\ \therefore T = 648{\rm{ N}}\end{array}\]

Hence tension in the string is 648 N.

Note: Travelling waves move from place to place whereas standing waves or stationary waves only occur for certain frequencies. In a transverse wave , positions of maximum and minimum amplitude travel through the medium.