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 ?

Options

A. `(1)/sqrt(2)cm`
B. `sqrt(2) cm`
C. `sqrt(3)/(2) cm`
D. `(2)/sqrt(3) cm`

Answer

a

Explanation

Mass per unit length of the string is
`mu=Ad=(0.80 mm^(2))xx(12.5 g//cm^(3))=0.01 kg//m`
`=(0.80xx10^(-6)m^(2))xx(12.5xx10^(3))=0.01 kg//m`
speed of transverse waves produced in the string
`v=sqrt((T)/(mu))=sqrt((64)/(0.01 kg//m))=80 m//s`
the amplitude of the source is `a=1.0 cm` and the frequency is `f=20 Hz`. the angular frequency is `omega=2pif=40 pi//s`.
also at `t=0`, the displacement is equal to its amplitute, `i.e.`, at `t=0, y=a`. the equation of motion of the source is, therefore,
`y=(1.0 cm)cos[(40pis^(-1))t]` ...`(1)`
the equation of the wave travelling on the string along the positive `X-`axis is obtained by replacing `t` by `[t-(x//v)]` in Eq. (i). It is, therefore,
`y=(1.0 cm)cos[940pis^(-1)){t-(x//v)}x]`
`=(1.0 cm)cos[(40pis^(-1))t-{(pi//2)m^-1)}x]`
the displacement of the particle at `x=50 cm` at time `t=0.05 s` is obtained from Eq. (ii).
`y=(1.0 cm)cos[(40pis^(-1))(0.05s)`
`-{(pi//2)m^(-1}(0.5 m)]`
`=(1.0 cm)cos[2pi-(pi//4)]`
`=1.0 cm//sqrt(2)=0.71 cm`
the velocity of the particle at position `x` at time `t` is also obtained from Eq. (ii).
`V=(dely)/(delt)=-(1.0 cm)(40pi s^(-1))t-{(pi//2)m^(-1)}x]`
`=-(40pi(cm)/(s))sin(2pi-(pi)/(4))`
`=-(40pi)/sqrt(2) cm//s=-89 cm//s`