Question

A man generates a symmetrical plus 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 $3m/s$ on the string & his hands passes $6$ times in eacgh seconds 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 D None

Firstly, we have to draw the pulse wave (see attached diagram).

If hand passes 6 times from the mean position in one second, then we know that string creates 3 wave lengths (λ) or 3 cycles after 1 second.

That means frequency (f) of the wave is $3Hz.$

Now we can use below equation to calculate the value of wave length.

$V=f\lambda$ (V = velocity of the wave)

$\lambda =\frac{V}{f}$

$=\frac{\left(3m/s\right)}{3}=1m$

If $\lambda =1m$, the point which have 3m distance is located at no.(6) (in the diagram).

to reach its upper extreme ----> have to travel 3λ/4 distance

time to travel $3\lambda =1$ second

time to travel $\lambda =\frac{1}{3}$ seconds

time to travel $\frac{3\lambda }{4}=\left(\frac{1}{3}\right)\times \left(\frac{3}{4}\right)$ seconds

$=\frac{1}{4}$ seconds $=0.25$ seconds