(a) Show that for a wave on a string, the kinetic energy per unit length of string is
Where /L is the mass per unit length.
(b) Calculate Uk(X, t) for a sinusoidal wave given by Eq. (15.7).
(c) There is also elastic potential energy in the string, associated with the work required to deform and stretch the string. Consider a short segment of string at position x that bas unstretched length fu, as in Fig. 15.13. Ignoring the (small) curvature of the segment, its slope is ∂y(x, t) ∂x. Assume that the displacement of the string from equilibrium is small, so that ∂y/∂x has a magnitude much less than unity. Show that the stretched length of the segment is approximately
Δx [1 + ½ (∂y(x, t)/ ∂x)2]
(Hint: Use the relationship √≈1+u≈1 + 1/2u valid for[u] « 1.)
(d) The potential energy stored in the segment equals the work done by the string tension F (which acts along the string) to stretch the segment from its unstretched length ax to the length calculated in part (c). Calculate this work and show that the potential energy per unit length of string is up
(x, t)= ½ F (∂y(x, t)/∂x)2
(e) Calculate up(x, t) for a sinusoidal wave given by Eq. (15.7).
(f)Show that Uk(x, t) = up (x, t} for all x and t.
(g) Show y (x, t}, Uk(x, t}, and up (x, t} as functions of x for t = 0 in one graph with all three functions on the same axes. Explain why Uk and up are maximum where y is zero, and vice versa.
(h) Show that the instantaneous power in the wave, given by Eq. (15.22), is equal to the total energy per unit length multiplied by the wave speed u. Explain why this result is reasonable.