Question

Transverse waves on a string have wave speed 8 m/s, amplitude 0.07 m, and wavelength 0.32 m . The waves travel in the -x-direction, and at t=0 the x=0 end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at x=36 m and time t=0.15 s. (d) How much time must elapse from the instant in part (c) until the particle at x=0.36 m next has maximum upward displacement?

Short answer

a) The frequency is f=25 Hz , period is T =0.04 s, and wave number is $\mathrm{k}=19.6{\mathrm{m}}^{-1}$.

b) A wave function, $y(x,t)=0.07\cos 2\pi \left(\frac{x}{0.32}+\frac{t}{0.04}\right)$

The transverse displacement of a particle,$y(0.36,0.15)=0.0495m$

c) Time, $\Delta t=0.005s$

Step-by-step solution

General formulas

The wave function of a sinusoidal wave can be given by

$\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{A}\mathbf{}\mathbf{cos}\mathbf{2}\mathbf{\pi }\left(\frac{x}{\lambda }\pm \frac{t}{T}\right)$

Here, A is the amplitude, x is the displacement, T is the period, and t is the time.

The relation between wave number (k) and wavelength $\left(\lambda \right)$ is

$\mathit{k}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\lambda }}$

The relation between frequency (f) and time period is

$\mathbf{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{T}}$

The relation between the speed of periodic wave (v) , frequency and wavelength is

$\mathbf{v}\mathbf{=}\mathbf{f\lambda }$

(a) Calculate frequency, time period and wave number

Consider the given data as below.

The velocity, $v=8m/s$

The wavelength, $\lambda =0.32m$

$$\begin{gathered} v=f\lambda \\ f=\frac{v}{\lambda } \\ =\frac{8}{0.32} \\ =25\mathrm{Hz} \end{gathered}$$

Time period is the inverse of the frequency.

$\begin{array}{r}T=\frac{1}{f}\\ =\frac{1}{25}\\ =0.04\mathrm{s}\end{array}$

Wave number is given by

$$\begin{gathered} \mathrm{k}=\frac{2\pi }{\lambda } \\ \mathrm{k}=\frac{2\pi }{0.32} \\ =19.6{\mathrm{m}}^{-1} \end{gathered}$$

(b) Put all the values in the equation of wave function

Consider the given data as below.

The amplitude, A=0.07 m.

Wave function will be given as

$\begin{array}{r}y(x,t)=0.07\cos 2\pi \left(\frac{x}{0.32}\mp \frac{t}{0.04}\right)\\ =0.07\cos 2\pi \left(\frac{x}{0.32}+\frac{t}{0.04}\right)\end{array}$

(c) Find the transverse displacement of a particle.

From the given equation, you can find the transverse displacement of a particle.

$$\begin{gathered} y(0.36,0.15)=0.07\cos 2\pi \left(\frac{0.36}{0.32}+\frac{0.15}{0.04}\right) \\ =0.0495\mathrm{m} \end{gathered}$$

(d) Calculate the instant of maximum upward displacement

The maximum upward displacement will equal the amplitude.

$$\begin{gathered} 0.07=0.07\cos 2\pi \left(\frac{0.36}{0.32}+\frac{t}{0.04}\right) \\ \cos 2\pi \left(\frac{0.36}{0.32}+\frac{t}{0.04}\right)=1 \end{gathered}$$

The cosine becomes unity when the angle equals $\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{\pi }\mathbf{,}\mathbf{4}\mathbf{\pi }\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{=}\mathbf{2}\mathbf{n\pi }$, where n=0,1,2,... .

$\begin{array}{r}2\pi \left(\frac{0.36}{0.32}+\frac{t}{0.04}\right)=2n\pi \\ t+0.045=0.04n\\ t=0.04n-0.045\end{array}$

In the given equation, substitute t=0.15 s to find the value of n.

$$\begin{gathered} 0.15=0.04n-0.045 \\ n=4.87 \\ n\approx 5 \end{gathered}$$

From this value, we will get

$$\begin{gathered} t=0.04\times 5-0.045 \\ =0.155s \end{gathered}$$

So, the time elapsed will be

$$\begin{gathered} ∆t=0.155-0.15 \\ =0.005s \end{gathered}$$