InterviewSolution
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InterviewSolution
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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. |
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Answer» 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. |
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