Question

When a load of 5kg is suspended from a spring it extends by 5cm.If we disturb, it will oscillate with a time period of​

Answer 1

Answer:

Linear SHM : F = - ks α=−w

2

x

Therefore, w

2

=

m

k

T=2π

k

m

The spring constant can be found out by k=

2cm

1kg

=

0.02m

10N

=500N/m

T=2π

k

m

=2×3.14×

500

5

=0.628s