A periodic transverse wave is established on a string such that there are exactly two cycles traveling along a 3.0-m section of the string. The crests move at 20.0 m/s.
-How long does it take a particle at the top of a crest to reach the bottom of an adjacent trough?
A)0.018 s
B)0.038 s
C)0.075 s
D)0.150 s
E)0.30 s

Question

Quizplus · Accepted Answer

One cycle of the wave includes one crest and one trough. Thus, there are four crests (and four troughs) in two cycles. Each 3.0-m section of the string corresponds to one wavelength, and in this case, there are two wavelengths in the 3.0-m section. Therefore, the wavelength is λ = 3.0 m / 2 = 1.5 m. The speed of the wave is given as v = 20.0 m/s. The frequency of the wave can be found using the wave equation: v = f λ, so f = v / λ = 20.0 m/s / 1.5 m = 13.3 Hz. The period of the wave is T = 1/f = 1/13.3 s ≈ 0.075 s. When a particle is at the top of a crest, it is at its maximum displacement, and when it reaches the bottom of an adjacent trough, it has to travel a distance equal to one-half of the wavelength. Thus, the time it takes for the particle to reach the bottom of an adjacent trough is t = (1/2) λ / v = (1/2) (1.5 m) / (20.0 m/s) ≈ 0.038 s. Therefore, the correct answer is B.

A Periodic Transverse Wave Is Established on a String Such