A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has a length 4.8m and mass 0.06 kg. QR has length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the 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.
(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`