Question

In the given figure a string of linear mass density m and length L is stretched by a force $F=(F_0-kt)N$ where $F_0$ & k are constant and t is time t = 0 a pulse is generated at the end P of the string As the pulse reaches the point Q the force vanishes The value of K in the above equation is

• A. $\frac{2}{3L}\sqrt{\frac{F_0^3}{m}}$
• B. $\frac{2}{3L}\sqrt{\frac{F_0^3}{2m}}$
• C. $\frac{1}{3L}\sqrt{\frac{F_0^3}{m}}$
• D. $\frac{1}{3L}\sqrt{\frac{F_0^3}{2m}}$

Answer 1

The correct option is A $\frac{2}{3L}\sqrt{\frac{F_0^3}{m}}$

Tension in the string

$T=F_0-kt$

Let velocity of pluse in the string is V and time take by pulse to travel distance from P to Q is $t_0$

Then at time $t_0,F=0,0=F_0-k{t}_0$

Hence $t_0=\frac{F_0}{K}$

Now V = $=\sqrt{\frac{T}{m}}$

$\frac{dx}{dt}=\sqrt{\frac{F_{0}kt}{m}}$

$\Rightarrow {\int }_0^{L}dx=\frac{1}{\sqrt{m}}{\int }_0^{t_0}(F_0-kt{)}^{1/2}dt$

$\Rightarrow L=\frac{1}{\sqrt{m}}\times \frac{-2}{3k}{\left[\right(F_0-kt{)}^{3/2}]}_0^{t_0}$

$L=\frac{-2}{3k\sqrt{m}}[0-F_0^{3/2}]$

$K=\frac{2F_0^{3/2}}{3L\sqrt{m}},k=\frac{2}{3L}\sqrt{\frac{F_0^3}{m}}$