Question

A rightward moving wave $y(x, t)=A \sin (k x-\omega t+\phi)$ has initial condition $y=0$ at $x=0$ and $t=0$. Find the value of $x$ and $t$ at $\phi=(1 / 2) n \pi$ for which particle acceleration will have the same value as for initial condition and $\phi=2 n \pi$, $n=0,2 \ldots$, 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)

$y(x, t)=A \sin (k x-\omega t+\phi)$.
For initial condition,
$a_{p}=\frac{d^{2} y}{d t^{2}}=-\omega^{2} A \sin \phi($ for $x=0$ and $t=0$ ) $=0$ for $\phi=2 n \pi$
Now, $a_{p}=-\omega^{2} A \sin (k x-\omega t+(1 / 2) n \pi), n=0,2,4$
$=0$ 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.