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

(i) 0.14 s (ii) `A_(r)=1.5cm, A_(b)=2cm`

Explanation

Given that `m_1=0.06 kg,m_2=0.2kg`
Let m' be the mass per unit length then `m_1=0.0125,m_2=0.078125 kg//m`
Wire PQR is undera tension of `T_0=80N`.A sinusodial wave pulse is sent from P.
(i) `v_1=` velocity of wave on PQ `=sqrt((T_0)/(M_1))=sqrt((80)/(0.0125))=80 m//s`
`v_2=` velocity of wave of QR `=sqrt((T_0)/(M_2))=sqrt((80)/(0.078125))=32 m//s`
Total tine taken for wave pulse to each from P to R
`=(PQ)/(V_1)+(QR)/(V_2)=(4.8/80+2.56/32)s=0.06+0.08=0.14s`
(ii) Amplitude of reflected wave : `A_r=((V_2-V_1)/(V_2+V_1))A_i`
`=((32-80)/(32+80))3.5=-1.5 cm`
`A_r` is -ve,so reflected wave is inverted ,
Amplitude of transmitted wave: `A_t=((2v_2)/(v_2+v_1))A_i=((2xx32)/(32+80))3.5=2cm`