Question

A transverse sine wave with an amplitude of 2.50mm and a wavelength of 1.80m travels from left to right along a long,horizontal stretched string with a speed od36.0m/s.Take the origin at the left end of the undisturbed string.At time t=0 the left end of the string has its maximum upward displacement .
a) What are the frequency,angular frequency and wave number of wave?
b) What is the function y(x,t) that describes the wave?
c) What is y(t) for a particle at the left end of the string?
d) What is y(t) for a particle 1.35m to the right of the origin?
e) What is the maximum magnitude of transverse velocity of any particle of the string?
f) Find the transverse displacement and the transverse velocity of a particle 1.35m to the right of the origin at time t=0.0625s.

Answer 1

The amplitude of a transverse wave, (A) = 2.50 mm = 2.50 * 10^-3 m, the wave-length, (λ) = 1.80 m and the wave speed, (v) = 36.0 m/s.

(a) The frequency of the wave,(f) =v/λ = 20.0 Hz. Its angular frequency, (ω) = 2f = 126 rad/s. The wave number of the wave, (k) =2π/λ = 3.49 rad/m.

(b) Since the wave is traveling from left to right, the functionto describe the wave,

y(x, t) = (2.50* 10^-3 m) cos [(3.49 rad/m)x - (126 rad/s)t] .

(c) At the left end of the string, x = 0, therefore, y(0, t) = +(2.50 10^-3 m) cos [(126 rad/s)t].

(d) At x = +1:35 m, y(x = +1:35 m, t) =+(2:50 * 10^-3 m) cos [(3.49 rad/m)(1.35 m) - (126 rad/s)t] = +(2:50 10^-3 m) cos [(4.71rad) - (126 rad/s)t].

(e) The maximum magnitude of the transverse velocity of a particle, (v_y)max = ωA

= (126 rad/s)(2.50* 10^-3 m) = 0:315 m/s.

(f)The transverse displacement of a particle at (x=1.35 m, t=0.0625 s),

y(x = 1.35 m; t = 0.0625 s) = (2.50* 10-3 m) cos [(3.49 rad/m)(1.35 m) - (126 rad/s)(0.0625 s)]

= 2.50 * 10^-3 m:

The transverse velocity of a particle at (x=1.35 m, t=0.0625 s),

v_y(x, t) = ωA sin (kx -ωt) = 6.90 *10-3 m/s:

Answer 2

The transverse velocity of the partice at x=1.35 and t=0.0625 is actually zero according to Mastering Physics.