Question

Write the equation of the wave shown in figure, if its position is shown at $t=0.6\text{sec}$. The wavelength is $15\text{cm}$ and amplitude is $4\text{cm}$ and the crest $X$ was at $x=0$ at $t=0$.

• A. $y(x,t)=\left(2\text{cm}\right)\sin \left[\right(1.68\text{rad/s})t-(1{\text{cm}}^{-1}\left)x\right]$
• B. $y(x,t)=\sin \left[\right(3.2\text{rad/s})t-(0.42{\text{cm}}^{-1}\left)x\right]$
• C. $y(x,t)=\left(2\text{cm}\right)\cos \left[\right(3.2\text{rad/s})t-(0.42{\text{cm}}^{-1}\left)x\right]$
• D. $y(x,t)=\left(4\text{cm}\right)\cos \left[\right(1.68\text{rad/s})t-(0.42{\text{cm}}^{-1}\left)x\right]$

Answer 1

The correct option is D $y(x,t)=\left(4\text{cm}\right)\cos \left[\right(1.68\text{rad/s})t-(0.42{\text{cm}}^{-1}\left)x\right]$

Given that, amplitude $\left(A\right)=4\text{cm}$ and wavelength $\lambda =15\text{cm}$
$K=\frac{2\pi }{\lambda }=\frac{2\pi }{15}=0.42{\text{cm}}^{-1}$
The wave travelled a distance $2.4\text{m}$ in $0.6\text{sec}$. Hence speed of the wave
$v=\frac{2.4}{0.6}=4\text{cm/s}$
$\therefore$ angular frequency $\left(\omega \right)=v\times K$
$=4\times 0.42=1.68\text{rad/s}$.
Since the wave is travelling along positive x-direction and crest $X$ was at $x=0$ at $t=0$, so we can write the wave equation as
$y(x,t)=A\cos (Kx-\omega t)$
or $y(x,t)=A\cos (\omega t-Kx)$
[as $\cos (-\theta )=\cos \theta$]
Therefore , the desired equation is
$y(x,t)=\left(4\text{cm}\right)\cos \left[\right(1.68\text{rad/s})t-(0.42{\text{cm}}^{-1}\left)x\right]$