Question

A particle of mass 200 gram execute simple harmonic motion the restoring force is provided by a spring of spring constant 80 N/m. find the time period?

Answer 1

m = 200g = 0.2kg
Spring force = 80N/ m
T = 2 X pi X root under 0.2/80
T= 2pi X root 2/800
T = 2pi X 1/20
T = 0.31 sec

Answer 2

Dear Student,

◆ Answer -

T = 0.3142 s

● Explanation -

# Given -

m = 200 g = 0.2 kg

k = 80 N/m

# Solution -

Time period of particle executing S.H.M. is given by -

T = 2π√(m/k)

T = 2 × 3.142 × √(0.2 / 80)

T = 6.284 √(1/400)

T = 6.284 / 20

T = 0.3142 s

Hence, time period of particle executing S.H.M. is 0.3142 s.

Thanks dear. Hope this helps you..