A transverse harmonic wave of amplitude 0.01 m is genrated at one end (x=0) of a long horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time the displacement of the particle at x=0.1 m is -0.005 m and that of the particle at x=0.2 m is +0.005 m. calculate the wavelength and the wave velocity. obtain the equetion of the wave assuming that the wave is traveling along the + x-direction and that the end x=0 is at he equilibrium position at t=0.

Explanation

Since the wave is travelling along `+ x-`direction and the displacement of the end `x=0` is at time `t=0`, the general equation of this wave is
`y(x,t)=A sin{(2pi)/(lambda)(vt-x_(1))}`...`(i)`
where `A=0.01 m`
when `x=0.1 m, y=-0.005 m`
Putting all values, we get `-0.005=0.01 sin{(2pi)/(lambda)(vt-x_(1))}`
where `x_(1)=0.1 m`
or, `sin{(2pi)/(lambda)(vt-x_(1)}=-(1)/(2)`
`:.` Phase sintheta`_(1)=(2pi)/(lambda)(vt-x_(1))=(7pi)/(6)`....`(ii)`
When `x=0.2 m y=+0.005`. therefore, we have
`+0.005=0.01 sin{(2pi)/(lambda)(vt-x_(2))}`
where `x_(2)=0.2 m`
`theta_(2)=(2pi)/(lambda)(vt-x_(2))=(pi)/(6)` ......`(iii)`
From Eqs.(ii)and (iii)
`Deltatheta=theta_(1)-theta_(2)=pi`
now, `Deltatheta=(2pi)/(lambda)Deltax`
Thus, `pi=-(2pi)/(lambda)(x_(1)-xx_(2))=(-2pi)/(lambda)(0.1-0.2)`
or `lambda=0.2 m`
Now, frequency g og the wave `=`frequency of the tuning for `k=500 Hz`.
hence, wave velocity
`v=f lambda=500xx0.2=100 m//s`
substituting for `A`,lambda and `v` in eq. (i) we get
`y(x,t)=0.01 sin(10pi(100t-x))`
this is the lequation of the wave where y and x are in metres and t in seconds.