Question

Two long strings A and B, each having linear mass density $1.2\times {10}^{-2}kg{m}^{-1},$ are stretched by different tensions 4.8 N and 7.5 N 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 = 20 ms on string B. When and where will the pulse on B overtake that on A ?

Answer 1

$m=1.2\times {10}^{-2}kg/m,T=4.8N$

$\Rightarrow Va=\sqrt{\frac{\left(4.8\right)}{(\frac{1}{2}\times {10}^{-2})}}=20m/s$

$m=1.2\times {10}^{-2}kg/m$

$Tb=\frac{7.5}{(1.2\times {10}^{-2})}=25m/s$

$t_1=0$ in string A

$t_1=0$ + 20 ms

$=20\times {10}^{-3}=0.02sec$

In 0.02 sec A has travelled

$=20\times 0.02=0.4m$

Relative speed between A and B

= 25 - 20 = 5 m/s

Time taken for B to overtake A

$=\frac{s}{v}=\frac{0.4}{5}=0.08sec.$