Question

A transverse wave along a string is given by y=$2\sin \left(2\pi \left(3t-x\right)-\frac{\pi }{6}\right)$, where x and y are in cm and 't' is in second. The acceleration of a particle located at x=4cm at t =1s is

• A. $36{\pi }^2$
  
• B. $72{\pi }^2$
• C. $18{\pi }^2$
• D. $12{\pi }^2$

Answer 1

The correct option is A $36{\pi }^2$

Given, $y=2\sin (2\pi (3t-x)-\frac{\pi }{6})$

We have to find acceleration at $x=4cm\&t=1s$

We know that $v=\frac{\partial y}{\partial t}$

So,

$v=12\pi \cos (2\pi (3t-x)-\frac{\pi }{6})$

Also , $a=\frac{\partial v}{\partial t}$

$a=-72{\pi }^2\sin (2\pi (3t-x)-\frac{\pi }{6})$

Putting values of $x\&t$

$a=-72{\pi }^2\sin (2\pi (3-4)-\frac{\pi }{6})$

$a=-72{\pi }^2\sin (-2\pi -\frac{\pi }{6})=36{\pi }^2$

So, A is the correct answer.