A rod of length L and mass m pivoted about one end is oscillating in vertical plane. The period of oscillation for small amplitudes is

T = 2(L/3g)

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T = 2(3L/2g)

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T = 2(2L/3g)

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T = 2(L/6g)

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Solution

The correct option is C T = 2(2L/3g)
The torque equation about the point of suspension is I = Mg(PQ) = Mg(L/2) sin . ]
For small oscillations sin = and moment of inertia about the point of suspension is $I = \frac{M L^{2}}{3} .$ Substituting the values, we get $\frac{d^{2}}{d t^{2}} = - (3 g / 2 L)$ .
Thus, time period T = 2(2L/3g)

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A rod of length L and mass m pivoted about one end is oscillating in vertical plane. The period of oscillation for small amplitudes is

The correct option is C T = 2(2L/3g)The torque equation about the point of suspension is I = Mg(PQ) = Mg(L/2) sin . ]For small oscillations sin = and moment of inertia about the point of suspension is I=ML23. Substituting the values, we get d2dt2=−(3g/2L). Thus, time period T = 2(2L/3g)