Question

A block string system is shown in the figure. Mass of block $A,B$ and $C$ is $5kg,2kg$, and $2kg$ respectively. Mass of rod is given as $1kg$ and its length is $1m$. A wave transverse is transmitted in the rod in between the block $B$ and $C$ once in forward and once backward direction. Time for string wave to reach one end to other end in forward direction is $\frac{2}{\sqrt{z}}(\sqrt{x}-\sqrt{y})sec$. Find $(x+y+z)$

Answer 1

Answer: 10

Acceleration of system $a=5m/s$
In case of backward direction wave transmission:
Tension at point from $r$ distance from $C$ in the string $BCT=15-5r$
Wave velocity, $v=\sqrt{\frac{T}{\mu }}$
$\Rightarrow v=\sqrt{\frac{15-5r}{1}}$
$\Rightarrow \frac{\partial r}{\partial t}=\sqrt{15-5r}$
$\Rightarrow \underset{0}{\overset{1}{\int }}\frac{\partial r}{\sqrt{15-5r}}=\underset{0}{\overset{t_1}{\int }}\partial t$
$\Rightarrow t_1=\frac{2}{5}\left(15^{1/2}-10^{1/2}\right)$
$\Rightarrow t_1=\frac{2}{\sqrt{5}}(\sqrt{3}-\sqrt{2})s$
In cace of forward direction wave transmission:
Tension at point from r distance from $B$ in the string $BCT$ $=10+5r$
$\Rightarrow v=\sqrt{\frac{10+5r}{1}}$
$\Rightarrow \frac{\partial r}{\partial t}=\sqrt{10+5r}$
$\Rightarrow {\int }_0^1\frac{\partial r}{\sqrt{10+5r}}={\int }_0^{t_2}\partial t$
$\Rightarrow t_2=\frac{2}{5}\left(15^{1/2}-10^{1/2}\right)$
$\Rightarrow t_2=\frac{2}{\sqrt{5}}(\sqrt{3}-\sqrt{2})s$
$t_1=t_2$