Physics, asked by SiddhantSinha8311, 1 year ago

A particle of mass 200 gm executes s.H.M. The restoring force is provided by a spring of force constant 80 n/m. The time period of oscillations is

Answers

Answered by mariatahir06
4
Time period for Shm is
T=2pie under root m/k
So put the values in above formula
Convert mass into kg which is 0.2kg
Put the values in the above formula
T=2pie under root 0.2/80
Taking under root of 0.2/80 as it is present in the formula we get 1/20
T= 2 pie 1/20
T= pie/10 seconds
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