Question

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

• A. $1.25\text{sec}$
• B. $1\text{sec}$
• C. $\frac{13}{12}\text{sec}$
• D. $1.4\text{sec}$

Answer 1

The correct option is A $1.25\text{sec}$

One upward and successive downward movement of hard will generate a complete wave.
$\Rightarrow$ Frequency of wave is,
$f=\frac{\left(6/2\right)}{1}=3{\text{s}}^{-1}$
wave speed, $V=3\text{m/s}$
$\Rightarrow \lambda =V\times T$
$\therefore \lambda =3\times \frac{1}{3}=1\text{m}$
Time taken by wave pulse to reach the point at distance $3\text{m}$ is,
$t_1=\frac{3}{V}=1\text{s}$
Time taken by particle at $3\text{m}$ distance to reach its upper extreme position
$t_2=\frac{3T}{4}$
$(\because$ particle will initially move downwards as $V\to +ve)$


$V_p=-\text{slope}\times V$
$\Rightarrow V_p\to -ve$ (downwards)
Hence, time taken by particle to reach its upper extreme for 1st time:
$t=t_1+t_2$
$\Rightarrow t=1+\frac{1}{4}=1.25\text{s}$