A long string having a cross- sectional area `0.80 mm^2` and density `12.5 g cm^(-3)` is subjected to a tension of 64 N along the X-axis. One end of this string is attached to a vibrator moving in transverse direction at a frequency fo 20Hz. At t = 0. the source is at a maximujm displacement y= 1.0 cm. (a) Write the equation for the wave. (b) What is the displacement of the particle of the string at x = 50 cm at time t = 0.05 s ? (c) What is the velocity of this particle at this instant ?

Answer

A, B, C, D

Explanation

(a) `v-=sqrt((T)/(rhoS)`
`=sqrt((64)/(12.5xx10^(3)xx0.8xx10^(-6))`
`=80 m//s`
(b) `omega = 2pif = 2pi(20) = 40pi rad//s`
`k=(omega)/(v) = (40pi)/(80) =(pi)/(2)m^(-1) `
`:. y = (1.0 cm) cos[(40pis^(-1))t-((pi)/(2)m^(-1))x]`
(c) Substituting `x=0.5 m and t=0.05 s`, we get
`y=(1)/(sqrt2) cm`
(d) Particle velocity at time `t`,
`v_(P) = (dely)/(delt)`
`=-(40pi cm//s) sin [(40pis^(-1))t-((pi)/(2)m^(-1))x]`
Substituting `x=0.5 m` and `t= 0.05 s`, we get
`v_(P) =89cm//s`