Question

The displacement y of a particle in a medium can be expressed as \[y = {10^{ - 6}}\sin \left( {110t + 20x + \dfrac{\pi }{4}} \right)\] m, where t is in seconds and x is in metre. The speed of the wave is?
A. 2000 m/s
B. 5 m/s
C. 20 m/s
D. \[5\pi \] m/s

Answer 1

Hint: By comparing the given wave equation with the general wave equation of the displacement of particles we find the respective physical quantities like amplitude, angular frequency and wave number. Then using a formula for speed we calculate the wave speed.

Formula used:
\[v = \dfrac{\omega }{k}\]
where \[\omega \] is the angular frequency and k is the wave number.

Complete step by step solution:
The general equation of the displacement of a particle is given as,
\[y = A\sin \left( {\omega t + kx + \phi } \right)\]
Here, A is the amplitude of the motion, i.e. the maximum displacement of the particle from the mean position, \[\omega \] is the angular frequency, k is the wave number and \[\phi \] is the phase angle.

The phase angle at initial time is called the phase angle. The given equation of the displacement is,
\[y = {10^{ - 6}}\sin \left( {110t + 20x + \dfrac{\pi }{4}} \right)\]m
On comparing both the equations, we get the amplitude of the displacement of a particle is \[{10^{ - 6}}m\]. The angular frequency is 110 rad/s. The wave number is 20 per metre. And the phase angle of the displacement of a particle is \[\dfrac{\pi }{4}\] rad.

The speed of the wave is given as,
\[v = \dfrac{\omega }{k} \\ \]
Putting the values, we get the speed of the wave as,
\[v = \dfrac{{110}}{{20}}\,m/s \\ \]
\[\therefore v = 5.5\,m/s\]
Hence, the speed of the wave is 5.5 m/s. From the given options, the nearest possible value for the speed is 5 m/s.

Therefore, the correct option is B.

Note: We should be careful about the units of the quantities written in the expression of the phase. We need to change the standard unit if the units are given other than the standard unit.