Question

Two long strings A and B, each having linear mass density 1·2×10-2 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

Time Taken will be 0.08 sec

Explanation:

mA = $1.2 \times 10^-^2 kgm^-^1,$

TA = 4.8 N  

⇒ VA = $\sqrt (T /m)$

= $20 ms^-^1$ mB  

= $1.2 \times 10^-^2 kgm^-^1$,  

TB = 7.5 N

⇒ VB = $\frac {T}{m} = 25 ms^-^1$ t = 0 in string  

At 1 = 0 + 20 ms  

= $20 \times 10^-^3$ = 0.02 sec  

In 0.02 sec A has travelled $20 \times 0.02$ = 0.4 mt  

Relative speed between A and B,

= 25 – 20 = $5\ ms^-^1$

Time taken for B for overtake A = $sv^_^1$  

= $\frac {0.4}{5}$= 0.08 sec