Question

A tuning fork vibrates 512 oscillations per seconds. Find the time period of
the vibrations produced and its frequency.

Answer 1

Given:

A tuning fork vibrates 512 oscillations per seconds.

To find:

• Time period
• Frequency of tuning fork ?

Calculation:

• Time period of an oscillatory body is defined as the time taken to complete one full oscillation.

• Frequency of an oscillatory body is defined as the number of oscillations completed in one second.

Now, since the tuning fork is completing 512 oscillations in one second, it's frequency is 512 Hz.

$\sf \therefore \: frequency = 512 \: hz$

Now, time period :

$\therefore \sf \: T = \dfrac{1}{frequency}$

$\implies\sf \: T = \dfrac{1}{512}$

$\implies\sf \: T = 0.0019 \: sec$

So, time period is 0.0019 sec and frequency is 512 Hz.