Question

Two long strings A and B, each having linear mass density , 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

Given,
Linear density of each of two long strings A and B, m = $1.2\times {10}^{-2}\mathrm{kg}/\mathrm{m}$
String A is stretched by tension T_a= 4.8 N.
String B is stretched by tension T_b= 7.5 N.
Let v_a and v_b be the speeds of the waves in strings A and B.
Now,
$$\begin{gathered} v_{\mathrm{a}}=\sqrt{\frac{T_a}{m}} \\ \Rightarrow v_{\mathrm{a}}=\sqrt{\frac{\left(4.8\right)}{\left(1.2\times {10}^{-2}\right)}}=20\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{b}}=\sqrt{\frac{T_{\mathrm{b}}}{m}} \\ \Rightarrow {\mathrm{v}}_{\mathrm{b}}=\sqrt{\frac{7.5}{\left(1.2\times {10}^{-2}\right)}}=25\mathrm{m}/\mathrm{s} \\ {t}_1=0\mathrm{in}\mathrm{string}\mathrm{A} \\ {t}_2=0+20\mathrm{ms}=20\times {10}^{-3}=0.02\mathrm{s} \end{gathered}$$
Distance travelled by the wave in 0.02 s in string A:
s$=20\times 0.02=0.4\mathrm{m}$
Relative speed between the wave in string A and the wave in string B, v'$=25-20=5\mathrm{m}/\mathrm{s}$
Time taken by the wave in string B to overtake the wave in string A = Time taken by the wave in string B to cover 0.4 m
$t\text{'}=\frac{s}{v\text{'}}=\frac{0.4}{5}=0.08\mathrm{s}$