Please someone help me with this question. A second transverse wave T2, of amplitude A has the same waveform as wave T1 but lags behind T1 by a phase angle of 60°. The two waves T1 and T2 pass through the same point. (i) On Fig. 5.1, draw the variation with time t of the displacement x of the point in wave T2.
\[\phi = \frac{2\pi}{\lambda} x\] phi = phase difference x equals distance between the waves
Equation of the transverse wave is y= Asin(wt+phi) here, A is the amplitude, w is the angular frequency, t is the time and phi is the phase difference. since T1 is y=Asin(wt) therefore for T2 y= Asin(wt-60). now follow the formula by thushananth to find the path difference where,
i think u can directly find x which is the distance between the two waves |dw:1400085647178:dw| i found that x = 0.5 cm so
Join our real-time social learning platform and learn together with your friends!