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Question

A rod of mass M and Length L is hinged at its one end and carries a block of mass m at its lower end. A spring of force constant k1 is installed at distance a from the hinge and another of force constant k2 at a distance b as shown in the figure. If the whole arrangement rests on a smooth horizontal table top, the frequency of vibration is
294515_e3e7d0d2029f4d71b6ff10a29127d137.png

A
12π  k1a2+k2b2L2(m+M3)
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B
12πk2+k1M+m
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C
12π   k2+k1a2b24M3+m
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D
12π    k1+k2b2a243m+M
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Solution

The correct option is A 12π  k1a2+k2b2L2(m+M3)
A is the point where the rod is hinged.
Let θ be the angular displacement from mean position.
The extension of spring connected at point B is =bθ (θ is so small that we take it as linear displacement)
So restoring force at point B=k2bθ
Moment of this force at point A=k2bθ.b=k2b2θ (b is the distance of force from A)
The contraction of spring connected at point c is:
=aθ
So, restoring force is =k1aθ
and moment of force about A is k1aθ.a=k1a2θ (a is the distance from point A of restoring force)
So, total restoring force =(k1a2θ+k2b2θ)θ
Now the total moment of inertia of the system about point A is:
I= Moment of inertia about point A +Mooment of inertial of mass m about A
=ML23+ML2
So, the angular acceleration
d2θdt2=(k1a2+k2b2)θI=(k1a2+k2b2)θL2(M3+M)
d2θdt2+(k1a2+k2b2)L2(M3+M)θ=0
So angular frequency ω=  (k1a2+k2b2)L2(M3+M)
and frequency, f=ω2π=12π  (k1a2+k2b2)L2(M3+M)

941395_294515_ans_5d735ce7fbe94ddca5ed9469bfa9dae5.png

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