Question

A rightward moving wave $y(x,t)=A\sin (kx-\omega t+\varphi )$ has initial condition $y=0$ at $x=0$ and $t=0$. Find the value of $x$ and $t$ at $\varphi =(1/2)n\pi$ for which particle acceleration will have the same value as for initial condition and $\varphi =2n\pi$, $n=0,2\dots$, if $\lambda$ and $f$ are wave length and frequency respectively.

• A. $\lambda ,\frac{1}{f}$
• B. $\frac{\lambda }{4},\frac{2}{f}$
• C. $2\lambda ,\frac{1}{f}$
• D. Both(a) and (c)

Answer 1

Answer: (D)

$y(x,t)=A\sin (kx-\omega t+\varphi )$.
For initial condition,
$a_p=\frac{d^{2}y}{d{t}^2}=-{\omega }^{2}A\sin \varphi ($ for $x=0$ and $t=0$ ) $=0$ for $\varphi =2n\pi$
Now, $a_p=-{\omega }^{2}A\sin (kx-\omega t+(1/2)n\pi ),n=0,2,4$
$=0$ if $x=\lambda$ and $t=\frac{1}{f}$ $=0$
if $x=2\lambda$ and $t=\frac{1}{f}$
Because $a_p=-{\omega }^{2}A\sin \left[\frac{2\pi }{\lambda }\cdot \lambda -2\pi f\cdot \frac{1}{f}+\frac{1}{2}n\pi \right]=0$
So, $x=\lambda$ and $2\lambda$ and $t=\frac{1}{f}$ are the required solutions.