91Ó°ÊÓ

The wave function of a sinusoidal wave can be given by

$\mathbf{y}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{A}\mathbf{}\mathbf{cos}\mathbf{2}\mathbf{Ï€}\left(\frac{x}{\mathrm{λ}}Â\pm \frac{t}{T}\right)$

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 $\left(\mathrm{λ}\right)$ is

$\mathit{k}\mathbf{=}\frac{\mathbf{2}\mathbf{π}}{\mathbf{λ}}$

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

$\mathbf{v}\mathbf{=}\mathbf{´Úλ}$

Consider the given data as below.

The velocity, $v=8m/s$

The wavelength, $\mathrm{λ}=0.32m$

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

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

$$\begin{gathered} \mathrm{k}=\frac{2\mathrm{π}}{\mathrm{λ}} \\ \mathrm{k}=\frac{2\mathrm{π}}{0.32} \\ =19.6{\mathrm{m}}^{−1} \end{gathered}$$

Consider the given data as below.

The amplitude, A=0.07 m.

Wave function will be given as

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

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

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

The maximum upward displacement will equal the amplitude.

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

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

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

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

$$\begin{gathered} 0.15=0.04n−0.045 \\ n=4.87 \\ n≈5 \end{gathered}$$

From this value, we will get

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

So, the time elapsed will be

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