Question

The equation of a wave travelling on a string is

$y=\left(0·10\mathrm{mm}\right)\sin \left[\left(31·4{\mathrm{m}}^{-1}\right)x+\left(314{\mathrm{s}}^{-1}\right)t\right]$.

(a) In which direction does the wave travel? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string?

Answer 1

Given,
Equation of the wave, $y=\left(0.10\mathrm{mm}\right)\sin \left(31.4{\mathrm{m}}^{-1}\right)x+\left(314{\mathrm{s}}^{-1}\right)t$
The general equation is $y=A\sin \left\{\left(\frac{2\pi x}{\lambda }\right)+\omega t\right\}$.
From the above equation, we can conclude:

(a) The wave is travelling in the negative x-direction.

(b) $\frac{2\pi }{\lambda }=31.4{\mathrm{m}}^{-1}$
$$\begin{gathered} \Rightarrow \lambda =\frac{2\pi }{31.4}=0.2\mathrm{m}=20\mathrm{cm} \\ \mathrm{And}, \\ \omega =314{\mathrm{s}}^{-1} \\ \Rightarrow 2\pi f=314 \\ \Rightarrow f=\frac{314}{2\pi } \\ =\frac{314}{2\times 3.14} \\ =50{\mathrm{s}}^{-1}=50\mathrm{Hz} \end{gathered}$$

Wave speed:
$$\begin{gathered} \nu =\lambda f=20\times 50 \\ =1000\mathrm{cm}/\mathrm{s} \end{gathered}$$

(c) Maximum displacement, A = 0.10 mm
$$\begin{gathered} \mathrm{Maximum}\mathrm{velocity}=a\omega =0.1\times {10}^{-1}\times 314 \\ =3.14\mathrm{cm}/\mathrm{s} \end{gathered}$$