Pretend a system is having Transverse waves. And those transverse waves on a string have wave speed 8.00 m/s amplitude 0.0700m and wavelength 0.320m. The waves travel in the - x dierection, and at t=0 the x=0 end of the string has its maximum upward displacement

a) Find the frequency, period, aand wave number of these waves.

b) Write a wave function describing the wave

c) Find the transverse displacement of a particle at x=0.360m at time t=0.150s

d) How much time must elapse from the instant in part c) until the particle at x=0.360m next has maximum upward displacement?

Answer: a) 25 Hz, 0.04s, 19.64. b) y (x, 0) = 0.07 sin 19.64x. c) - 0.019 m. d) 0.045s

Explanation: wave speed (v) = 8m/s, amplitude (A) = 0.07m and wavelength (λ) = 0.32m

A)

Recall that v = fλ

8 = f (0.32)

f = 8 / 0.32 = 25 Hz.

But T = 1/f

T = 1/25 = 0.04s

Wave number (k) = 2π/λ = 2 (3.142) / 0.32 = 19.64

B)

y (x, t) = A sin (kx - wt) but t = 0

Hence, y (x, 0) = A sin kx

y (x, 0) = 0.07 sin 19.64x

C) recall that y (x, t) = A sin (kx - wt), we are to find y (x, t) when x = 0.360m and t = 0.150s

w=2πf = 2 (3.142) * 25 = 157.14 rad/s

A = 0.07m

k = 19.64

y (x, t) = 0.07 sin {19.65 (0.360) - 157.14 (0.15) }

y (x, t) = 0.07 sin { 7.074 - 23.571}

y (x, t) = 0.07 sin (-16.497)

y (x, t) = 0.07 * (-0.283)

y (x, t) = - 0.019 m

D) wave speed = 8m/s, x = 0.360 m

Wave speed = distance / time

8 = 0.360/t

t = 0.360/8 = 0.045s