A transverse harmonic wave on a string is described by y(x,t) =5.0 sin(48t + 0.0264 x + pi/6),where x and y are in cm and t in sec. The positive direction of x is from left to right. (a) What are its amplitude and frequency? (b) What is the least distance between two success in crests in the wave?

A transverse harmonic wave on a string is described by y(x,t) =5.0 sin

1.	A transverse harmonic wave on a string is described by y(x,t) =5.0 sin(48t + 0.0264 x + pi/6),where x and y are in cm and t in sec. The positive direction of x is from left to right. (a) What are its amplitude and frequency? (b) What is the least distance between two success in crests in the wave?
Answer» SOLUTION :Here, y(X,t) = `5.0 sin(48t + 0.0264x + pi/6)` The general EQUATION of a plane progressive wave is, `v(x,t) = a sin[((2pi)/lambda(vt +x) + phi)]` It is OBSERVED that the given equation represent a travellig waveform right to left. Velocity, `V=48/0.0264 = 1818.18 cm s^(-1), r=5` cm. (a) Amplitude and frequency: Amplitude, `(2pi)/lambda = 0.0264` or `lambda = (2pi)/0.0264 cm = (2 xx 3.14)/(0.264) = 6.28/0.0264 = 237.8` cm frequency, From the equation `v= lambda v` `v= v/lambda =(1818.18)/(2pi) xx 0.0264 =(1818.18)/(2 xx 3.14) xx 0.0264` `=(1818.18)/(6.28) xx 0.0264 = 289.51 xx 0.0264 = 7.64` Hz (b) To find least distance between two successive crests in the wave. `lambda =(2pi)/(0.0264) =(2 xx 3.14)/(0.0264) = 6.28/0.0264` =237.8 cm= 2.38 m (C ) When `x = lambda/4` `phi =(2pi)/lambda xx lambda/4 = pi/2` rad.