Question

Two tuning forks have frequencies 412 Hz and 409 Hz respectively.Calculate the period of the beat

Answer 1

The time period of the beat is 0.34 seconds.

Explanation:

The difference in frequency of two tuning forks is called beat frequency. Two tuning forks have frequencies 412 Hz and 409 Hz respectively. Their beat frequency is given by :

$f=f_2-f_1$

$f=412-409$

f = 3 Hz

If T is the period of the beat. It is given by :

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

$T=\dfrac{1}{3}$

T = 0.34 seconds

So, the time period of the beat is 0.34 seconds. Therefore, it is the required solution.

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