Question

A wave is represented by the equation
$y=\left(0·001\mathrm{mm}\right)\sin \left[\left(50{\mathrm{s}}^{-1}\right)t+\left(2·0{\mathrm{m}}^{-1}\right)x\right]$.

(a) The wave velocity = 100 m s⁻¹.
(b) The wavelength = 2⋅0 m.
(c) The frequency = 25/π Hz.
(d) The amplitude = 0⋅001 mm.

Answer 1

(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm

$y=\left(0·001\mathrm{mm}\right)\sin \left[\left(50{\mathrm{s}}^{-1}\right)t+\left(2·0{\mathrm{m}}^{-1}\right)x\right]$
Equating the above equation with the general equation, we get:
$$\begin{gathered} y=A\sin \left(\omega t-kx\right) \\ \omega =\frac{2\mathrm{\pi }}{T}=2\pi v \\ k=\frac{2\mathrm{\pi }}{\mathrm{\lambda }} \end{gathered}$$
Here, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.
$$\begin{gathered} A=0.001\mathrm{mm} \\ \mathrm{Now}, \\ 50=2\pi \nu \\ \Rightarrow \mathrm{\nu }=\frac{25}{\mathrm{\pi }}\mathrm{Hz} \end{gathered}$$