Question

The equation of a travelling sound wave along $x-axis$ is given as $S=9\sin (400t-2.7x)$, where $S$ is measured in ${10}^{-4}\text{m}$, $t$ in seconds and $x$ in metres. Then:

• A. Ratio of displacement amplitude to wavelength of wave is $3.86\times {10}^{-4}$.
• B. The velocity amplitude of the particle is $36\times {10}^{-2}\text{m/s}$.
• C. Ratio of the velocity amplitude of the particle to the wave speed is $2.43\times {10}^{-3}$.
• D. The velocity amplitude of the particle is $3.6\times {10}^{-2}\text{m/s}$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Ratio of displacement amplitude to wavelength of wave is $3.86\times {10}^{-4}$.
B The velocity amplitude of the particle is $36\times {10}^{-2}\text{m/s}$.
C Ratio of the velocity amplitude of the particle to the wave speed is $2.43\times {10}^{-3}$.

Given equation is :
$S=9\sin (400t-2.7x)$
So ,
$S_0=9\times {10}^{-4}\text{m}$
$\omega =400\text{rad/s}$
$k=2.7\text{rad/m}=\frac{2\pi }{\lambda }$
$\Rightarrow \lambda =\frac{2\pi }{2.7}\text{m}$
So, the ratio of displacement amplitude of the particle to the wavelength of the wave $=\frac{S_o}{\lambda }=\frac{9\times {10}^{-4}}{2\pi /2.7}=3.86\times {10}^{-4}$
Velocity amplitude of the particle :
($V_p{)}_{max}=A\omega =S_o\omega$
$=9\times {10}^{-4}\times 400\text{m/s}$
$=36\times {10}^{-2}\text{m/s}$
and wave speed $\left(V_w\right)=f\lambda$
$=\left(\frac{\omega }{2\pi }\right)\times \left(\frac{2\pi }{k}\right)=\frac{\omega }{k}$
$=\frac{400}{2.7}\text{m/s}=148.15\text{m/s}$
So ,
$\frac{(V_p{)}_{max}}{V_w}=\frac{36\times {10}^{-2}}{148.15}=2.43\times {10}^{-3}$
Option (a), (b), (c) correct.