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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

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