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Question

Find the period of small oscillations of the bob of mass m shown in the figure. Given, mass of the rod is also m.


A
T=2π2m3k
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B
T=2πm3k
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C
T=2π7m3k
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D
T=2π4m3k
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Solution

The correct option is D T=2π4m3k
Suppose the small mass m is pulled down by a small distance x and released.


From conservation of energy principle, we obtain
12kx2mgx+12mv2+12lω2=constant.....(1)
where v is speed of the mass m when the elongation of the spring is x & ω is the angular speed of the rod.
Neglecting loss of gravitational P.E term,
12kx2+12mv2+12lω2=constant.....(2)
As x is very small, x=lθ2.
v=dxdt=lω2
Putting this value in (2) we obtain,
12k(l2θ)2+12m(l2.ω)2+12ml212ω2=constant
Differentiating w.r.t time, we obtain
12kl24.2θ.dθdt+12m.l242ωdωdt+12ml212.2ωdωdt=0.
dωdt=kl24⎢ ⎢ ⎢1ml24+ml212⎥ ⎥ ⎥.θ
dωdt=3k4mθ
Since dωdt=α.
α=3k4m.θ
Compare with α=ω2oscθ.
Therefore the angular frequency of oscillation is given by, ωosc=3k4m
T=2π4m3k is the time period of osillations.

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