Question

a particle of mass 200gm executes SHM.the restoring force is provided by a spring of force constant 80N/m.the time period of oscillation is?​

Answer 1

time period of oscillation is 0.31 seconds

Step-by-step explanation:

we shall know the formula:

`T = 2\pi \sqrt{(m)/(k) }` ......where m is mass and k is force constant.

change 200gm into kg, then it will be 0.2 kg

T = 2π√0.2÷80

T = 0.314 seconds

T = 0.31 seconds rounded to 2 decimal places.