Question

A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is $350m{s}^{-1}$. (a) By how much is the phase changed at a given point in 2.5 ms? (b) What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation?

Answer 1

Here given, n =100, v =350 m/s

$\Rightarrow \lambda =\left(\frac{v}{n}\right)=\frac{350}{100}=3.5m$

In 2.5 ms, the distance travelled by the particle is given by

$Dx=(350\times 2.5\times {10}^{-3})$

So, phase difference,

$\varphi =\frac{2\pi }{\lambda }\times Dx$

$=\left(\frac{2\pi \times 350\times 2.5\times {10}^{-3}}{3.5}\right)$

$=\left(\frac{\pi }{2}\right)$

(b)In the second case,

Given $\Delta \eta =10cm={10}^{-1}m$

So, $\varphi =\frac{2\pi }{x}\Delta x$

$=\frac{2\pi \times {10}^{-1}}{\left(\frac{350}{100}\right)}=\frac{2\pi }{35}$