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Question

A rigid rod of mass m, with a ball of mass M attached to the free end is restrained to oscillate in a vertical plane as shown in the figure. Find the natural frequency of oscillation.


A
f=12π3k27M+7m
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B
f=1π3k27m+7m
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C
f=12πk27M+7m
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D
f=1πk27M+7m
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Solution

The correct option is A f=12π3k27M+7m
At equilibrium position, deformation of the spring be x0.


Using principle of moments about point O we get ,
kx0L4=Mg(3L4)+mgL4 .....(1)
When the rod is rotated clockwise through an angle θ from equilibrium position, the spring will be compressed further by x.
Then, restoring torque about point O
τ=[k(x+x0)L4cosθMg3L4 cos θmgL4cosθ]
Using (1) in the above equation, we get
τ=[kx(L4)]cosθ
For small values of θ, cosθ1
τ=kL4x
But we know that, τ=Iα and x=L4θ
Iα=kL216θ ........(2)
Moment of inertia about point O is given by
I=IMO+ImO
I=M(34L)2+mL212+m(L4)2=916ML2+748mL2
Substituting this in (2) we get,
α=3k27M+7mθ
Comparing this with equation of Angular SHM α=ω2θ we get,
ω=3k27M+7m
Since, f=ω2π
f=12π3k27M+7m
Thus, option (a) is the correct answer.

Alternate solution :

Let the suppose the spring is stretched by x units, causing the system to make an angle θ about point O.
For small values of θ we can say that, x=L4θ.
Restoring force in the spring F=kx=kL4θ
Restoring torque about point O is given by
τ=F×L4=kL216θ
But we know that,
τ=Iα
Moment of inertia about point O is given by
I=IMO+ImO
I=M(34L)2+mL212+m(L4)2=916ML2+748mL2
Using this we get,
α=3k27M+7mθ
Comparing this with α=ω2θ and using f=ω2π we get,
f=12π3k27M+7m
Thus, option (a) is the correct answer.

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