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?

Answer 1

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 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,

\[v_a  = \sqrt{\frac{T_a}{m}}\] 

\[ \Rightarrow  v_a  = \sqrt{\frac{\left( 4 . 8 \right)}{\left( 1 . 2 \times {10}^{- 2} \right)}} = 20  m/s\] 

\[ v_b  = \sqrt{\frac{T_b}{m}}\] 

\[ \Rightarrow  v_b  = \sqrt{\frac{7 . 5}{\left( 1 . 2 \times {10}^{- 2} \right)}} = 25  m/s\] 

\[ t_1  = 0  \text{ in  string  A }\] 

\[ t_2  = 0 + 20  ms = 20 \times  {10}^{- 3}  = 0 . 02  s\]
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
\[t' = \frac{s}{v'} = \frac{0 . 4}{5} = 0 . 08  s\]