Question

In the figure shown strings $AB$ and $BC$ have masses $m$ and $2m$ respectively. Both are of same length $l$. Mass of each string is uniformly distributed on its length. The string is suspended vertically from the ceiling of a room. A small jerk wave pulse is given at the end ' $C$. It goes up to upper end ' $A$ ' in time ${}^{\prime }{t}^{\prime }$. If $m=2kg,l=\frac{9610}{1681}m,g=10m/s^2,\sqrt{2}=1.4,\sqrt{3}=1.7$ then ${}^{\prime }t$ ' is equal to :

• A. $\frac{620}{697}s$
• B. $\frac{434}{205}s$
• C. 2 s
• D. None of these

Answer 1

Answer: (C)

We use $v_{BC}=\sqrt{gx}$
$\underset{0}{\overset{l}{\int }}\frac{dx}{\sqrt{x}}=\underset{0}{\overset{t_{BC}}{\int }}\sqrt{g}dt$
$2\sqrt{l}=g{t}_{BC}$
$t_{BC}=2\sqrt{\frac{l}{g}}$
$v_{AB}=\sqrt{\frac{\left(2m+\frac{m}{l}x\right)g}{\frac{m}{l}}}$
$=\sqrt{\left(2+\frac{x}{l}\right)gl}$
$\underset{0}{\overset{l}{\int }}\frac{dx}{\sqrt{2+\frac{x}{l}}}=\underset{0}{\overset{t_{AB}}{\int }}\sqrt{gl}dt$
$2l\sqrt{2+\frac{x}{l}}=\sqrt{gl}{t}_{AB}$
$2l(\sqrt{3}-\sqrt{2})=\sqrt{gl}{t}_{AB}$
$t_{AB}=2(\sqrt{3}-\sqrt{2})\sqrt{\frac{l}{g}}$
$t=t_{BC}+t_{AB}=2\sqrt{\frac{l}{g}}(1+\sqrt{3}-\sqrt{2})$
$=1.3\times 2\sqrt{\frac{961}{1181}}=2.6\times \frac{31}{41}$
$=1.96s\simeq 2s$