,

The displacement is a maximum when the term in parentheses in the denominator is zero; the denominator is the sum of two squares and is minimized when x = vt, and the maximum displacement is A. At x = 4.50 cm, the displacement is a maximum at t = (4.50 × 10 –2 m)/(20.0 m/s) = 2.25 × 10 –3 s. The displacement will be half of the maximum when (x – vt) 2 = A 2 , or t = (x ± A)/v = 1.75 × 10 –3 s and 2.75 × 10 –3 s.

Of the many ways to obtain the result, the method presented saves some algebra and minor calculus, relying on the chain rule for partial derivatives. Specifically, let u = u(x, t) = x – vt, so that if f(x, t) = g(u), = = and = = – v.

(In this form it may be seen that any function of this form satisfies the wave equation; In this case, y(x, t) = A 3 (A 2 + u 2 ) –1 , and so

= , = –

= v , = – v 2 ,

and so the given form for y(x, t) is a solution to the wave equation with speed v.