Question

The speed of wave of frequency 500 Hz is 360 m/s. the minimum distance between two particles on it having a phase difference of 60° is .......... (a) 0.23 m (b) 0.12 m (c) 8.33 m (d) 60 m

Answer 1

Answer:

(b) 0.12 m

Explanation:

As per the question,

Given speed of the wave 'v' = 360 m/s

Frequency of the wave 'n' = 500 Hz

Phase difference 'x' = 60° = (π/3) rad

So,

Wavelength of the wave is given by formula:

$Wavelength = \frac{Speed}{Frequency}$

On putting the values of given data, we get

$\lambda = \frac{v}{n}$

$\lambda= \frac{360}{500}$

∴ λ = Wavelength = 0.72 m

Formula used for phase difference is given by:

$\phi =\frac{2 \pi x}{\lambda}$

where,

Ф = phase difference

x = path difference = the minimum distance between two particles

$\Rightarrow \frac{\pi}{3} =\frac{2 \pi x}{\lambda}$

$\Rightarrow x =\phi \times \frac{\lambda}{2\pi}$

$\Rightarrow x = \frac{\pi}{3} \times \frac{0.72}{2\pi}$

∴ x = 0.12 m

Therefore, the minimum distance between two particles = x = 0.12 m

Hence the correct option is (b).