The incident wave in problem 1 [y = 4 sin (3p t - 6p x)] and the reflected wave [y = -4 sin (3p t + 6p x)] produce a standing wave.

(b)(iii) Sketch the standing wave at t = (3/15)T

y = -8 cos3p t sin6p x

[Remember from Q1 that T = 2/3 s]
at t = (3/15)T = (3/15)(2/3) = 2/15 s
y = -8 cos3p (2/15) sin6p x
= -8 cos0.4p sin6p x
= -8 cos 72° sin 6p x
= -2.47 sin 6p x

This is just a negative sine curve with a maximum of 2.47 and a wave length of 1/3
(Remember k = 6p = 2p /l)

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