Question

a sinusoidal transverse wave travels along a long, stretched string. the amplitude of this wave is 0.0817 m, its frequency is 2.51 hz, and its wavelength is 1.75 m.
what is the shortest transverse distance d between a maximum and a minimum of the wave?
how much time δt is required for 52.3 cycles of the wave to pass a stationary observer?
viewing the whole wave at any instant, how many cycles n are there in a 36.5 m length of string?

Explanation:

Step1: Find the shortest transverse distance \(d\)

The amplitude \(A = 0.0817\space m\). The shortest transverse distance between a maximum and a minimum of a wave is \(2A\).
\(d=2\times0.0817\)

Step2: Find the time \(\Delta t\) for \(n = 52.3\) cycles

The frequency \(f=2.51\space Hz\). The period \(T=\frac{1}{f}\). The time for \(n\) cycles is \(\Delta t=nT=\frac{n}{f}\).
\(\Delta t=\frac{52.3}{2.51}\)

Step3: Find the number of cycles \(N\) in \(L = 36.5\space m\)

The wavelength \(\lambda=1.75\space m\). The number of cycles \(N=\frac{L}{\lambda}\).
\(N=\frac{36.5}{1.75}\)

Answer:

\(d = 0.1634\space m\)
\(\Delta t=20.836653\space s\)
\(N = 20.857143\space cycles\)