Question

The pulse shown is figure has a speed $8\text{cm/s}$. The linear mass density of right string is 4 times that of the left string. What is ratio of the heights of the transmitted pulse and the reflected pulse?
[Assume tension to be same]

• A. 4
• B. 1
• C. 2
• D. 6

Answer 1

The correct option is C 2

Assuming tension be the same in both string
So, $T_1=T_2=T$
Linear mass density of left string $={\mu }_1$
Linear mass density of right string = ${\mu }_2$
From question , ${\mu }_2=4{\mu }_1$
As we know wave speed $v=\sqrt{\frac{T}{\mu }}$
$v\propto \frac{1}{\sqrt{\mu }}$ [since tension in both strings are same]
$\Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{{\mu }_2}{{\mu }_1}}=\sqrt{\frac{4{\mu }_1}{{\mu }_1}}=2$
$\Rightarrow v_2=\frac{v_1}{2}=\frac{8}{2}=4\text{cm/s}[\because v_1=8\text{cm/s (given)}]$
As from formula,
$A_r=\left(\frac{v_2-v_1}{v_2+v_1}\right)A_i----\left(1\right)$
$\&A_t=\left(\frac{2v_2}{v_2+v_1}\right)A_i----\left(2\right)$
Dividing eq (2) by (1),
$\frac{A_t}{A_r}=\left(\frac{2v_2}{v_2-v_1}\right)=\frac{2\times 4}{4-8}=\frac{8}{-4}=-2$
Taking magnitude, $\left|\frac{A_t}{A_r}\right|=2$