Question

For the wave shown in the figure, write the equation of the wave if its position is shown at $t=0$ speed of wave is 200 m/s.

• A. $y(x,t)=0.12\sin \pi [x-2000t]\text{m}$
• B. $y(x,t)=0.12\sin [31.4x-6200t]\text{m}$
• C. $y(x,t)=0.12\sin [3.14x-6283t]\text{m}$
• D. $y(x,t)=0.12\sin [31.4x-6283t]\text{m}$

Answer 1

The correct option is D $y(x,t)=0.12\sin [31.4x-6283t]\text{m}$

From the figure, we can deduce that,
Amplitude $\left(A\right)=0.12m$
$\frac{3}{2}\lambda =0.3$
$\Rightarrow \lambda =\frac{2\times 0.3}{3}=0.2\text{m}$
frequency $\left(f\right)=\frac{v}{\lambda }=\frac{200}{0.2}=1000\text{Hz}$
Angular wavenumber $\left(k\right)=\frac{2\pi }{\lambda }=\frac{2\pi }{0.2}=31.4{\text{m}}^{-1}$
and $\omega =2\pi f=2\pi \times 1000\approx 6283\text{rad/s}$

At $t=0,x=0,\frac{dy}{dx}=+ve$
Therefore given curve is a sine curve
Hence, equation of wave travelling in positive x-direction should have the form

$y(x,t)=A\sin (kx-\omega t)$
Substituting the obtained values, we get
$y(x,t)=0.12\sin \left[\right(31.4x-6283t]\text{m}$
Thus, option (d) is the correct answer.