Question

The equation of a traveling sound wave is$y=6.0\sin (600t-1.8x)$ where $y$ is measured in ${10}^{-5}m$, $t$ in second and $x$in meter.

(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave.

(b) Find the ratio of the velocity amplitude of the particles to the wave speed.

Answer 1

Step 1: Given data:

The equation of a traveling sound wave is$y=6.0\sin (600t-1.8x)\dots \dots \dots \dots \dots \dots \left(1\right)$

Where $y$ is measured in${10}^{-5}m$

Step 2: Formula used:

In the standard wave equation:

$y\left(x,t\right)=A\sin \left(kx-\omega t+\varphi \right)\dots \dots \dots \dots \dots \left(2\right)$

Where$y$ is the vertical position $x$ is the horizontal position variable, and $t$ is the time variable,

Suppose that $V_y$ the velocity amplitude of the wave.

Velocity $v=\frac{dy}{dt}$

Time period $T=\frac{2\pi }{\omega }$

Wave speed $v=\frac{\lambda }{T}$

Part (a):

Step 3: Calculating the wavelength of the wave:

Comparing equation (1) and (2) with the standard wave equation-

We get,

$\omega =600\mathrm{rad}/\mathrm{s}$

The formula of $K=\frac{2\pi }{\lambda }whereKiswavecons\tan t$

Therefore,

$\begin{array}{rcl}600& =& \frac{2\pi }{\lambda }\\ \lambda & =& \frac{2\pi }{1.8}=0.00103\end{array}$

Step 4: Calculate the ratio of the displacement amplitude of the particles to the wavelength of the wave:

After comparing the given equation with the standard waveform-

$A=6\times {10}^{-5}m$

$\begin{array}{rcl}\frac{A}{\lambda }& =& \frac{6.0\times 1.8\times {10}^{-5}m}{2\mathrm{\pi }}\\ \frac{A}{\lambda }& =& 1.7\times {10}^{-5}m\end{array}$

Therefore the ratio of the displacement amplitude of the particles to the wavelength of the wave$\begin{array}{l}1.7\times {10}^{-5}m\end{array}$

Part (b):

Step 5: Calculating the velocity amplitude of the particles:

Using formula,

$v=\frac{dy}{dt}$

$$\begin{gathered} v=\frac{d\left[6\sin \left(600t-1.8x\right)\right]}{dt} \\ v=3600\cos \left(600t-1.8x\right)\times {10}^{-5}m/s \end{gathered}$$

Therefore $V_y=3600\times {10}^{-5}m/s$

Step 6: Now calculate the time period:

Using the formula time period can be calculated as follows-

$$\begin{gathered} T=\frac{2\pi }{\omega } \\ T=\frac{2\pi }{600} \end{gathered}$$

Step 7: Calculate the wave velocity:

Wave velocity can be calculated as-

$$\begin{gathered} v=\frac{600}{1.8} \\ v=\frac{100}{3}m/s \end{gathered}$$

Step 8: Calculate the required ratio:

The ratio of the velocity amplitude of the particles to the wave speed

$$\begin{gathered} \left(\frac{V_y}{v}\right)=\frac{3600\times 3\times {10}^{-5}}{1000} \\ \left(\frac{V_y}{v}\right)=1.1\times {10}^{-4} \end{gathered}$$

Therefore the ratio of the velocity amplitude of the particles to the wave speed is $1.1\times {10}^{-4}$