Question

The displacement equation of a travelling sound wave along $x$ axis is given by $s=5\times {10}^{-5}\sin (500t-2x)$, where $s$ and $x$ are in meters and $t$ in seconds. Choose the correct option among the following.

• A. Ratio of displacement amplitude of the particles to the wavelength of wave is $1.1\times {10}^{-5}$.
• B. Ratio of displacement amplitude of the particles to the wavelength of wave is $1.59\times {10}^{-5}$.
• C. Ratio of the velocity amplitude of the particles to the wave speed is $2\times {10}^{-4}$.
• D. None of these.

Answer 1

The correct option is B Ratio of displacement amplitude of the particles to the wavelength of wave is $1.59\times {10}^{-5}$.

Given that,
$s=5\times {10}^{-5}\sin (500t-2x)$

We have the genreal equation of a wave as

$y=A\sin (\omega t-kx)$

Comparing the above equations gives

$s_0=5\times {10}^{-5}\text{m}$
(Amplitude of dispalcement)

$\omega =500{\text{s}}^{-1}$

$k=2{\text{m}}^{-1}$

Now,
$\frac{s_0}{\lambda }=\frac{s_0}{\left(\frac{2\pi }{k}\right)}=\frac{s_{0}k}{2\pi }$

$=\frac{5\times {10}^{-5}\times 2}{2\pi }$

$=1.59\times {10}^{-5}$

Ratio of displacement amplitude of the particles to the wavelength of wave is $1.59\times {10}^{-5}$. So, option $\left(b\right)$ is correct.

Wave velocity $v=\frac{\omega }{k}=\frac{500}{2}\text{m/s}$

Velocity of the particle $\left(v_p\right)$ is given by

$\frac{ds}{dt}=\frac{d}{dt}(5\times {10}^{-5}\sin (500t-2x\left)\right)$

$\Rightarrow v_p=5\times {10}^{-5}\times 500\cos (500t-2x)$

Ratio of the velocity amplitude of the particles to the wave speed is

$\therefore \frac{(v_p{)}_{max}}{v_{wave}}=\frac{5\times 500\times {10}^{-5}}{\frac{500}{2}}$

$=2\times 5\times {10}^{-5}$

$={10}^{-4}$

So option $\left(c\right)$ is incorrect.