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. Calculate,
(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 joint `Q`.

Answer

A, B, C, D

Explanation

Tension, `T = 80N`
Amplitude of incident wave,` A_i = 3.5 cm `
Mass per unit length of wire PQ is
`m_1 = 0.06/4.8 = 1/80 kg//m`
and mass per unit length of wire QR is
`m_2 = 0.2/2.56 = 1/12.8 kg//m`

(a) Speed of wave in wire PQ is
`v_1 = sqrt(T//m_1) = sqrt(80/(1//80)) = 80 m//s`
and speed of wave in wire QR is
`v_2 = sqrt(T//m_2)`
`= sqrt (80/(1//12.8)) = 32 m//s`
`:.` Time taken by the wave pulse to reach from
P to R is
`t = 4.8/v_1 + 2.56/v_2 = (4.8/80 + 2.56/32)s `
`= 0.14s ` .
(b) The expressions for reflected and transmitted
amplitudes `(A_r and A_t)` in terms of `v_1, v_2 and A_i` are as follows:
`A_r = (v_2-v_1)/(v_2 +v_1)A_i `
and `A_t = ((2v_2)/(v_1+v_2))A_i`
Substituting the values, we get
`A_r = ((32 - 80)/(32+80)) (3.5) = -1.5 cm `
i.e. the amplitude of reflected wave will be
1.5 cm. Negative sign of `A_r` indicates that there
will be a phase change of `pi` in reflected wave.
Similarly, `A_t = ((2xx32)/(32+80)) (3.5) = 2.0 cm`
i.e. the amplitude of transmitted wave will be
2.0 cm.
The expressions of `A_r and A_t` are derived as
below.
Derivation
Suppose the incident wave of amplitude `A-i` and
angular frequency `omega` is travelling in positive
x-direction with velocity `v_1`, then we cna write
`y_i = A_i sin omega [t-x//v_1]`.........(i)
In reflected as well as transmitted wave, `omega` will not
change, therefore, we can write.
`y_r = A_r sin omega [t+ x//v_1]`......(ii)
and `y_t = A_t sin omega [t - x//v_2]......(iii)`
Now as wave is continuous, so at the boundary
`(x=0)`.
Continuity of displacement requires
`y_i + y_r = y_i for x=0`
Substituting from Eqs. (i), (ii) and (iii) in the
above, we get
`A_i +A_r = A_t` ........(iv)
Also at the boundary, slope of wave will be
continuous, i.e.
`(dely_i)/(deltax) + (dely_r)/(deltax) = (dely_t)/(deltax)` [for x=0]
which gives `A_i-A_r = (v_1/v_2)A_t` ..........(v)
Solving Eqs. (iv) and (v) for `A_r and A_t, ` we get the required equations, i.e.
`A_r = (v_2-v_1)/(v_2+v_1) A_i and A_t = (2v_2)/(v_2+v_1) A_i` .