Question

A block string system is shown in the figure. Mass of block $A, B$ and $C$ is $5\, kg , 2\, kg$, and $2\, kg$ respectively. Mass of rod is given as $1 \,kg$ and its length is $1 \,m$. 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

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