Question

show that bigilar oscillation are SHM derive the expression for periodic time of bifilar oscillation with parallel threads​

Answer 1

Explanation:

Time period T= time required to complete one oscillation.

At equilibrium T

0

=mg.

For small displacement θ

Restoring force =−mgsinθ

for small θ sinθ≈θ

⇒fe=−mgθ=−mg(

l

x

)

acceleration a=

m

fe

=

l

−g

.x

We know for SHM, a=−w

2

x

⇒ On comparing we get w=

l

g

∴ Time period of oscillation T=

w

2Π

=2Π

g

l

.