Transverse waves on a string have wave speed 8.00 m/s, amplitude 0.0900 m, and w

Question

Transverse waves on a string have wave speed 8.00 m/s, amplitude 0.0900 m, and wavelength 0.305 m. The waves travel in the -x-direction, and at t = 0 the x = 0 end of the string has its maximum upward displacement. Find the following properties of these waves, frequency f = Hz period T = s wave number k = rad/m Write a wave function describing the wave. (Use the following as necessary: t and x.) y(x, f) = m Find the transverse displacement of a particle at x = 0.360 m at time t = 0.150 s. cm How much time must elapse from the instant in part (c) until the particle at x = 0.360 m next has maximum upward displacement s

Explanation / Answer

given

wave speed, v = 8 m/s

Amplitude, A = 0.09 m

wavelength, lamda = 0.305 m

a) use the relation, v = lamda*f

==> f = v/lamda

= 8/0.305

= 26.23 m/s

Time period, T = 1/f

= 1/26.23

= 0.038 s

k = 2*pi/lamda

= 2*pi/0.305

= 20.6 rad/s

b) angular frequncy, w = 2*pi*f

= 2*pi*26.23

= 164.8 rad/s

y(x,t) = A*cos(k*x - w*t)

= 0.09*cos(20.6*x - 164.8*t)

c) at x = 0.36 m, t = 0.15 s

y(x,t) = 0.09*cos(20.6*0.36 - 164.8*0.15)

= 0.00227 m

= 0.227 cm

d) let time t y = 0.09

0.09 = 0.09*cos(20.6*0.36 - 164.8*t)

1 = cos(20.6*0.36 - 164.8*t)

0 = 20.6*0.36 - 164.8*t

==> t = 20.6*0.36/164.8

= 0.045 s

so, at time = T - t

it comes to maximum upward displacement

time = 2*0.038 - 0.045

= 0.031 s

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