Question

The velocity of sound in sea water is about $\mathbf{1530}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{sec}$. Write an equation for a sinusoidal sound wave in the ocean, of amplitude 1 and frequency$\mathbf{1000}\mathbf{}\mathbf{hertz}$ .

Short answer

The equation for a sinusoidal sound wave is $y=\sin 2000\pi \left(\frac{x}{1530}-t\right)$.

Step-by-step solution

Given data

The dimensions are given as follows:

Amplitude,$\mathrm{A}=1.$

Frequency, $\mathrm{f}=1000\mathrm{hertz}.$

Velocity,$\mathrm{v}=1530\mathrm{m}/\sec .$

Concept of Sinusoidal sound wave

The trigonometric cosine function can be used to explain a sinusoid's unique functional form, and the most fundamental sinusoid can be written as$\mathit{A}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{}\mathbf{2}\mathit{\pi }\mathbf{(}\mathit{f}\mathit{t}\mathbf{+}\mathit{\varphi }\mathbf{\left)}\mathbf{\right)}$

Calculation to find the dimension of the wave equation

Following is a standard wave equation, $y=A\sin \left(\frac{2\pi }{\lambda }(x-vt)\right)$.

Time period is the time required to travel a distance of .

So, $\left(dis\tan ce\right)=\left(speed\right)\left(time\right).$

$\lambda =vT$

Since, $T=\frac{1}{f}.$

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

Hence, $v=\lambda f$.

Put the value of the obtained dimensions in the sinusoidal wave equation

Now, solve the wave equation as follows:

$$\begin{gathered} y=A\sin \left(\frac{2\pi }{\lambda }(x-vt)\right) \\ y=A\sin \left(\frac{2\pi v}{\lambda }\left(\frac{x}{v}-t\right)\right) \end{gathered}$$

Since,$(v=\lambda f).$ .

$$\begin{gathered} y=A\sin \left(\frac{2\pi \left(\lambda f\right)}{\lambda }\left(\frac{x}{v}-t\right)\right) \\ y=A\sin 2\pi f\left(\frac{x}{v}-t\right) \end{gathered}$$

Now, substitute all the value in above equation as follows:

$y=\sin 2000\pi \left(\frac{x}{1530}-t\right)$