Question

A transverse wave along a string is given by $y=2sin \left(2 \pi \left(3 t - x\right) + \frac{\pi }{4}\right),$ where, $x$ and $y$ are in $cm$ and $t$ in second. Find the acceleration of a particle located at $x=4 \, cm$ at $t=1 \, s$ .

• A. $36\sqrt{2} \, \pi ^{2} \, cms^{- 2}$
• B. $36\pi ^{2} \, cms^{- 2}$
• C. $-36\sqrt{2} \, \pi ^{2} \, cm \, s^{- 2}$
• D. $-36\pi ^{2} \, cms^{- 2}$

Answer 1

Answer: (C)

We know that $v=\frac{d y}{d t}$
$=\frac{d}{d t} \, 2sin \left[2 \pi \left(3 t - x\right) + \frac{\pi }{4}\right]$
$a=\frac{d v}{d t}=\frac{d}{d t}12\pi cos \left(2 \pi \left(3 t - x\right) + \frac{\pi }{4}\right)$
$a=-72 \pi^2 \sin \left(2 \pi(3 t-x)+\frac{\pi}{4}\right)$
at $t=1$ and $x=4 \, cm$
$a=-36 \, \sqrt{2}\pi ^{2} \, cm \, s^{- 2}$