Question

Two long strings A and B, each having linear mass density $12\times {10}^{-2}kg{m}^{-1}$, are stretched by different tensions $48N$and $75N$ respectively and are kept parallel to each other with their left ends at $x=0$. Wave pulses are produced on the strings at the left ends at $t=0$ on string A and at $t=20ms$ on string B. When and where will the pulse on B overtake that on A?

Answer 1

Step 1: Given data

Linear mass density of string A and B, $\mu =1.2\times {10}^{-2}kg{m}^{-1}$
Tension on string A, $T_A=4.8N$
Tension on string B, $T_B=7.5N$

Wave starts in A at $t=0$ and in B at $t=20ms=0.02s$

Step 2: Formula Used

Velocity of wave in string is given as,
$V=\sqrt{\frac{T}{\mu }}$
where $T$ is the tension on the string and $m$ is the mass

Step 3: Calculate velocity of waves

In string A,
$\begin{array}{rcl}{v}_A& =& \sqrt{\frac{48}{12\times {10}^{-2}}}\\ & =& 20m/s\end{array}$

In string B,
$\begin{array}{rcl}{v}_B& =& \sqrt{\frac{75}{12\times {10}^{-2}}}\\ & =& 25m/s\end{array}$

Formula for velocity is,
$v=\frac{x}{t}$
where $x$ is displacement and $t$ is time elapsed.

Step 4: Calculate time required to overtake

Time elapsed between start of wave in A and B is,
$t_e=0.02-0=0.02s$

Displacement of wave in A when $t=0.02s$ is,

$\begin{array}{rcl}x& =& vt\\ & =& 20\times 0.02\\ & =& 0.4m\end{array}$

The relative speed between A and B is $25-20=5m{s}^{-1}$

Time taken for B to reach $x=0.4m$ is the same time required by it to overtake A. So B overtakes A in,

$\begin{array}{rcl}t& =& \frac{x}{v}\\ & =& \frac{0.4}{5}\\ & =& 0.08s\end{array}$

Therefore, B overtakes A in $0.08s$ when $x=0.4m$.