Question

The wave described by \[y = 0.25\sin \left( {10\pi x - 2\pi t} \right)\], where x and y are in metres and t in seconds, is a wave travelling along the
A. +ve x direction with frequency 1Hz and wavelength \[\lambda = 0.2\]m
B. -ve x direction with amplitude 0.25 m and wavelength \[\lambda = 0.2\]m
C. –ve x direction with frequency 1 Hz
D. +ve x direction with frequency p Hz and wavelength \[\lambda = 0.2\]m

Answer 1

Hint:The general equation of the wave displacement is given as \[y = A\sin \left( {kx - \omega t} \right)\]. By comparing the given equation with the general equation of the wave displacement we can find the wavelength, frequency and the amplitude.

Formula used:
\[k = \dfrac{{2\pi }}{\lambda }\]
where k is the wave number of the wave with wavelength \[\lambda \].

Complete step by step solution:
The given displacement equation of the wave is,
\[y = 0.25\sin \left( {10\pi x - 2\pi t} \right)\]
The general equation of the wave displacement is given as,
\[y = A\sin \left( {kx - \omega t} \right)\]
On comparing both the equations, we get
The amplitude of the displacement,
\[{y_{\max }} = 0.25m\],

The wave number of the wave,
\[k = \dfrac{{2\pi }}{\lambda }\],
\[\Rightarrow 10\pi = \dfrac{{2\pi }}{\lambda }\]
\[\Rightarrow \lambda = 0.2m\]
So, the wavelength is 0.2 m
\[\omega = 2\pi \nu \]
\[\Rightarrow 2\pi = 2\pi \nu \]
\[\Rightarrow \nu = 1\,Hz\]
So, the frequency of the wave is 1 Hz.

The direction of the motion of the wave can be determined using phase expression of the wave equation. If the wave is travelling towards +x then with time the phase must increase or otherwise it will decrease. The phase of the wave is \[\left( {10\pi x - 2\pi t} \right)\].

The phase of the wave equation is pure constant, so the derivative of the phase with respect to time will be zero.
\[\dfrac{d}{{dt}}\left( {10\pi x - 2\pi t} \right) = 0 \\ \]
\[\Rightarrow 10\pi \dfrac{{dx}}{{dt}} - 2\pi \dfrac{{dt}}{{dt}} = 0 \\ \]
\[\Rightarrow 10\pi \dfrac{{dx}}{{dt}} - 2\pi = 0 \\ \]
\[\Rightarrow 10\pi v - 2\pi = 0 \\ \]
\[\Rightarrow v = \dfrac{{2\pi }}{{10\pi }}m/s \\ \]
\[\therefore v = 0.2\,m/s\]
As the velocity is positive, the wave is travelling towards +x direction. Hence, the wave is travelling along +ve x direction with frequency 1Hz and wavelength \[\lambda = 0.2\] m.

Therefore, the correct option is A.

Note: The phase of the wave equation is pure constant, i.e. dimensionless, so the overall derivative of the phase is zero that does not mean that phase of the wave remains constant. It changes with x and time.