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Question

A rigid uniform rod AB of length L and mass m is hinged at C such that AC=L/3, CB=3L/3. Ends A and B are supported by springs of spring constant k. The natural frequency of the system is given by


A
k2m
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B
km
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C
2km
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D
5km
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Solution

The correct option is D 5km
Method I:



θ=yAL3=yB2L3

I0=IC+m(2L3L2)2

IC=mL212

I0=mL212+mL236=mL29

Taking, ΣM0=0

(kyA×L3)+(kyB×2L3)+I0¨θ=0

kθ(L3)2+kθ(2L3)2+mL29¨θ=0

(mL29)¨θ+(5kL29)θ=0

¨θ+(5km)θ=0

ω2n=5km;ωn=5km

Method II :



Restoring torque:

T=(kL3θ)L3+2L3(k2L3θ)

Icα=(5kL29)θ

α=(5kL29)θmL12+m(L6)2=(5km)θ

We know that,

ωn=Angular accelerationAngular displacement

=  (5km)θθ=5km

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