Question

Two long strings $\text{A}$ and $\text{B}$ each having linear mass density $1.2\times {10}^{-2}\text{kg/m}$ are stretched by different tensions $4.8\text{N}$ and $7.5\text{N}$ respectively and are kept parallel to each other with their left ends at $x=0$. Wave pulses are produced on the string at the left ends at $t=0$ on string$\text{A}$ and $t=20\text{ms}$ on string $\text{B}$. When will the pulse on $\text{B}$ overtake that on $\text{A}$ for the first time?

• A. $0.5\text{s}$
• B. $0.25\text{s}$
• C. $0.1\text{s}$
• D. $0.4\text{s}$

Answer 1

The correct option is C $0.1\text{s}$

Given that
$T_A=4.8\text{N}$ $T_B=7.5\text{N}$
${\mu }_A=1.2\times {10}^{-2}\text{kg/m}$ ${\mu }_B=1.2\times {10}^{-2}\text{kg/m}$
So
$v_A=\sqrt{\frac{T_A}{{\mu }_A}}=\sqrt{\frac{4.8}{1.2\times {10}^{-2}}}=20\text{m/s}$
$v_B=\sqrt{\frac{T_B}{{\mu }_B}}=\sqrt{\frac{7.5}{1.2\times {10}^{-2}}}=25\text{m/s}$
The pulse produced on $\text{A}$ and $\text{B}$ meets when, $x_A=x_B$

Let ${}^{\prime }{t}^{\prime }$ be the time at which both pulses meet
So,
$x_A=x_B\Rightarrow 20t=25(t-0.02)$
$\Rightarrow 5t=0.5=0.1\text{s}$

Thus, option (c) is the correct answer.