Question

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

Answer 1

Answer:

512 oscillations=1 sec

therefore, i oscillation=1/512 sec

=0.0019 sec

time period= 0.0019 sec

frequency = 512 oscillations

Answer 2

Given : A tuning fork vibrates 512 oscillations per seconds.

To Find : the time period of the vibrations produced and its frequency.​

Solution:

Frequency of an oscillatory body is the number of oscillations completed in one second

tuning fork vibrates 512 oscillations per seconds.

Hence Frequency = 512 Hz

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

512 oscillations   in 1 sec

=> 1 oscillations   in 1/512 sec

=> 1 oscillations   in 0.00195 sec

Time period = 1.95 ms

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time period of the vibrations produced and its frequency.

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