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?

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

2b2t

explanation:

2b2t