Question

A wave has a frequency of 0.5 kHz and two particles with a phase difference of \pi /3 are 1.5 cm apart. Calculate: the time period of the wave.

Answer 1

Answer:

The time period of the wave is 0.002 sec.

Explanation:

Given that,

Frequency = 0.5 kHz

Phase difference $\phi=\dfrac{\pi}{3}$

Path difference = 1.5 cm

We need to calculate the time period

Using formula of time period

The frequency is the reciprocal of time period.

$f =\dfrac{1}{T}$

$T=\dfrac{1}{f}$

Where, f = frequency

Put the value into the formula

$T=\dfrac{1}{0.5\times10^{3}}$

$T=0.002\ sec$

Hence, The time period of the wave is 0.002 sec.