A transverse displacement wave is traveling cm a string It i

A transverse displacement wave is traveling cm a string. It is described by y(x. t) - A sin(kx - omega t + 3 pi/4). Nearby, a thin, uniform rod of mass m is hung from one end and allowed (oscillate (through small angles only) about that end as a physical pendulum, Find expressions (using only known values) for each of the following: Find (he minimum s(right-line distance between two point(s on the string whose acceleration magnitudes are each 2/3 of their maximum magnitudes (at the same moment). If the pendulum has a frequency (Hz,) that is 1/l00th of the wave\'s frequency, find the length of the rod. Assume the following known values: A, k, co, m. g

Solution

a) y(x,t) = Asin(kx - wt + 3pi/4)

v = d(y(x.t)) / dt = -Aw cos(kx - wt + 3pi/4)

a = dv / dt = -Aw^2 sin(kx - wt + 3pi/4)

a _max = Aw^2

|a| = Aw^2 sin(kx - wt + 3pi/4)

at t= 0,

a(x, 0 ) = Aw^2 sin(kx - 3pi/4)

2(Aw^2 ) / 3 = Aw^2 sin(kx - 3pi/4)

sin(kx - 3pi/4) = 2/3

kx - 3pi/4 = - 0.232pi, 0.232pi , 0.7677pi, 1.232pi, 1.7677pi , .......

kx = 0.518pi, 0.982pi, 1.518pi, 1.982pi, 2.518pi,.......

x = 0.518pi/k, 0.982pi/k , 1.518pi/k, 1.982pi/k, 2.518pi/k,.......

difference between two consecutive x\'s.

deltax = 0.464pi/k, 0.532pi/k, 0.464pi/k, 0.532pi/k, .....

shortest distance = 0.0464pi/k

b) frquency of wave = w / 2pi

frequency of pendulum = (w/2pi) / 100 = w /200pi

frequency of pendulum of mass m and length L = sqrt(3g / 2L) / 2pi

so sqrt(3g / 2L) / 2pi = w /200pi

sqrt(3g / 2L) = w / 100

3g / 2L = w^2 / 10000

L = 30000g / 2w^2 = 15000g / w^2