The time period of oscillation of a simple pendulum of length l is T1 and the time period of oscillation of a uniform rod of the same length l pivoted at one end and oscillating in a vertical plane is T2 . Amplitude of oscillations in both the cases is small. Then T1/T2 is:

Question

The time period of oscillation of a simple pendulum of length l is T1 and the time period of oscillation of a uniform rod of the same length l pivoted at one end and oscillating in a vertical plane is T2 . Amplitude of oscillations in both the cases is small. Then T1/T2 is: (A) (1/√3) (B) 1 (C) √(4/3) (D) √(3/2)

Tardigrade Admin · Accepted Answer

Time period of simple pendulum is given by

T_{1} = 2 π \frac{l}{g}

and time period of uniform rod in given position is given by

T_{2} = 2 π \frac{I n er t ia f a c t or}{s p r in g f a c t or}

Here, inertia factor=moment of inertia of rod at one end

= \frac{m l ^{2}}{12} + \frac{m l ^{2}}{4} = \frac{m l ^{2}}{3}

Spring factor=restoring torque per unit angular displacement

= m g \times \frac{l}{2} \frac{s in θ}{θ}

= m g \times \frac{l}{2}

(if

θ

is small)

∴

T_{2} = 2 π \frac{m l ^{2} /3}{m g l /2} = 2 π \frac{2}{3} \frac{l}{g}

Hence,

\frac{T _{1}}{T _{2}} = \frac{3}{2}

$\frac{1}{3}$

$1$

$\frac{4}{3}$

$\frac{3}{2}$

Q. The time period of oscillation of a simple pendulum of length l is T1​ and the time period of oscillation of a uniform rod of the same length l pivoted at one end and oscillating in a vertical plane is T2​ . Amplitude of oscillations in both the cases is small. Then T1​/T2​ is:

3​1​

1

34​​

23​​

$\frac{1}{3}$

$1$

$\frac{4}{3}$

$\frac{3}{2}$