A long wire `PQR` is made by joining two wires `PQ` and `QR` of equal radii `PQ` has length `4.8 m` and mass `0.06 kg`. `QR` has length `2.56m` and mass `0.2 kg`. The wire `PQR` is under a tension of `80N`. A sinusoidal wave-pulse of amplitude `3.5cm` is sent along the wire `PQ` from end `P`. No power is dissipated during the propagation of the wave-pulse. caculate,
(a) the time taken by the wave-pulse to reach the other end `R` of the wire, and
(b) the amplitude of the reflected and transmitted wave-pulse after the incident wave-pulse crosses the jjoint `Q`.

Answer

A, B, C, D

Explanation

(a) Mass per unit length of `PQ`
`m_(1) = (0.06)/(4.8)kg//m`
. ^(P) `0.06 kg` 0.2kg /`4.8 m`

Mass per unit length of `QR`, `m_(2) = (0.2)/(2.56)kg//m`
Velocity of wave on `PQ` is
`v_(1) = sqrt((T)/(m_(2))) = sqrt((80)/(0.06//4.8)) = 80ms^(-1)` [`because` `T = 80 N` given]
Velocity of wave in `QR` is
`v_(2) = sqrt((T)/(m_(2))) = sqrt((80)/(0.2//2.56)) = 32m//s`
Timetaken for the wave to reach from `P to R`
`= t_(PQ)+ t_(QR)`
`= (4.8)/(80) + (2.56)/(32) = 0.14 s`
(b) When the wave which initiates from `P` reaches `Q` (a denser medium) then it is partly reflected and partly transmitted.
In this case the amplitude of reflected wave
`A_(r) = ((v_(2) - v_(1))/(v_(2) - v_(1)))A_(i)` ...(i)
where `A_(i) = amplitude of incident wave`.
Also amplitude of transmitted wave is
`A_(t) = ((2v_(2)/(v_(1) + v_(2)))A_(i)` ...(ii)
From (i), (ii)
Therefore, `A_(t) = 2 cm and A_(r) = -1.5 cm`.