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Question
Two long strings A and B, each having linear mass density
\[1 \cdot 2 \times {10}^{- 2} kg m^{- 1}\] , 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?
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Solution
Given,
Linear density of each of two long strings A and B, m =\[1 . 2 \times {10}^{- 2} kg/m\]
String A is stretched by tension Ta= 4.8 N.
String B is stretched by tension Tb= 7.5 N.
Let va and vb be the speeds of the waves in strings A and B.
Now,
\[v_a = \sqrt{\frac{T_a}{m}}\]
\[ \Rightarrow v_a = \sqrt{\frac{\left( 4 . 8 \right)}{\left( 1 . 2 \times {10}^{- 2} \right)}} = 20 m/s\]
\[ v_b = \sqrt{\frac{T_b}{m}}\]
\[ \Rightarrow v_b = \sqrt{\frac{7 . 5}{\left( 1 . 2 \times {10}^{- 2} \right)}} = 25 m/s\]
\[ t_1 = 0 \text{ in string A }\]
\[ t_2 = 0 + 20 ms = 20 \times {10}^{- 3} = 0 . 02 s\]
Distance travelled by the wave in 0.02 s in string A:
s
\[= 20 \times 0 . 02 = 0 . 4 m\]
Relative speed between the wave in string A and the wave in string B, v'
\[= 25 - 20 = 5 m/s\]
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
\[t' = \frac{s}{v'} = \frac{0 . 4}{5} = 0 . 08 s\]
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