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
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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
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