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 ?

Explanation

(a) The mass of 1m long part of the string is
`m=(0.80mm^2)xx (1m)xx(12.5g cm^(-3))`
`=(0.80xx 10^(-6)m^2)xx(12.5xx10^3kgm^(-3))`
=0.01kg.
The linear mass density is `mu = 0.01 kg m^(-1).` The wave speed is `upsilon = sqrtF//mu`
`=sqrt((64N)/(0.01kgm^(-1)) = 80m s^(-1)` (b) The amplitude of the soure is A = 1. cm and the frequency is v= 20Hz. The angular frequency is `omega 2piv = 40pi s^(-1)` Also at t= 0, the displacement is equal to its amplitude i.e., at t = 0x A. The equation of motion of the source is, therefore, `y = (1.0cm) cos [(40pis^(-1))t] ... (i)`
The equation of the wave travelling on the sstring along the positive X-axis is obtained by replacing t with `t-x//upsilon` in equation (i) it is, therefore,
`y= (1.0cm) cos [40 pis^(-1)(t-(x)/(upsilon))]`
`=(1.0 cm)cos[(40 pi s^(-1))t-((pi)/(2)m^(-1))x], ....(ii)`
Where the value of `upsilon` has been put from part (a). (c ) The displacement of the particle at x=50 cm at time t = 0.05 s is by equation (ii), `y=(1.0cm) cos[(40pis^(-1)) (0.05s) - ((pi)/(2)m^(-1)) (0.5m)]`
`=(1.0cm) cos [2pi - (pi)/(4)]`
`(1.0cm)/(sqrt2) = 0.71cm.` (d) The velocity of the particle at position x at time t is, by equation (ii), `upsilon = dely)/(delt) =-(1.0) (40pi s^(-1)) sin[(40pi s^(-1) t-((pi)/(2)m^(-1)x]`
Putting the values of x and t, `upsilon = -(40pi cm s^(-1)) sin ((2pi - (pi)/(4))`
`=(40pi)/(sqrt2)cm s^(-1)~~ 89 cm s^(-1).`