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

The time period of the wave is 0.002 sec.

Step-by-step explanation:

Given that,

Frequency = 0.5 kHz

Phase difference
`\phi=(\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 =(1)/(T)`

`T=(1)/(f)`

Where, f = frequency

Put the value into the formula

`T=(1)/(0.5*10^(3))`

`T=0.002\ sec`

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