Question

A sound wave of frequency $245Hz$ travels with a speed of $300m{s}^{-1}$ along the positive $x-axis$. Each point of the medium moves to and fro through a total distance of $6cm$. What will be the mathematical expression of the travelling wave?

• A. $Y(x,t)=0.03[\sin 5.1x–(0.2\times {10}^3\left)t\right]$
• B. $Y(x,t)=0.06[\sin 5.1x–(1.5\times {10}^3\left)t\right]$
• C. $Y(x,t)=0.06[\sin 0.8x–(0.5\times {10}^3\left)t\right]$
• D. $Y(x,t)=0.03[\sin 5.1x–(1.5\times {10}^3\left)t\right]$

Answer 1

The correct option is D

$Y(x,t)=0.03[\sin 5.1x–(1.5\times {10}^3\left)t\right]$

Step 1: Given data

Frequency, $f=245Hz$

Speed of sound, $v=300m{s}^{-1}$

Distance, $$\begin{gathered} d=6cm \\ d=0.06m \end{gathered}$$

Step 2: Find the mathematical expression of the travelling wave

General expression is given by

$Y=A\sin (kx–\omega t)$

Amplitude is given by

$$\begin{gathered} A=d/2 \\ A=0.03m \end{gathered}$$

Angular frequency is given by

$$\begin{gathered} \omega =2\pi f \\ \omega =2\pi \times 245 \\ \omega =1.54\times {10}^{3}rad{s}^{-1} \end{gathered}$$

Propagation constant is

$$\begin{gathered} k=\omega /v \\ k=(1.54\times {10}^3)/300 \\ k=5.1m^{-1} \end{gathered}$$

The general expression became,

$Y(x,t)=0.03[\sin 5.1x–(1.5\times {10}^3\left)t\right]$

Hence, option D is correct.