Question

A man generates a symmetrical pulse in a string by moving his hand up and down.At t = 0 the point in his hand moves downward. the pulse travels with speed of 3 m/s on the string & his hands passes 6 times in each second from the mean position . then the point on the string at a distance 3m will reach its upper extreme first time at time t =

• A. $1.25sec.$
• B. $1sec$
• C. $\frac{13}{12}\sec$
• D. none

Answer 1

The correct option is B $1sec$

$f=3Hz$

$\omega =2\pi f=6\pi rad/sec$

$k=\frac{\omega }{v}=\frac{6\pi }{3}=2\pi rad/sec$

$y=Asin(kx-\omega t)$

$=Asin(2\pi x-6\pi t)$

putting $y=\pm A$ and $x=3$, we get

$6\pi -6\pi t=\frac{\pi }{2}$

$6\pi t=\frac{11}{2}\pi$

$t=\frac{11}{12}s$