Question

A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum. How does it differ from the time period calculated using the formula for a simple pendulum?

Answer 1

Answer ⇒ Time period of physical pendulum is 0.90 seconds and difference is 0.01 seconds.

Explanation ⇒

We know that the time period of a physical pendulum is given by,

$T = 2\pi \sqrt{\frac{i}{mgl} }$, where i is the moment of inertia about the support.  

We know that the M.I. about the centre of hollow sphere = 2/3 mr² Thu, by parallel axis theorem, it can be obtained that,

Moment of inertia about the support about the support,

i = 2mr²/3 + ml²

∴ i = m(2r²/3 + l²)

∴ i = m(2(0.02)²/3 + (0.20²))  

[∵ r = 0.02 cm and l = 0.18 + 0.02 = 0.20 cm.]

∴ i = 0.040m

∴ T = 2π√(0.040m/m × 9.8× 0.20)

∴ T = 0.90 s

Hence, the time period of oscillations = 0.90 seconds.

Now, For simple pendulum,

T' =2π√(l/g)

∴ T' = 2π√(0.20/9.8)

∴ T' = 0.89 s

Difference = T' - T

 = 0.90 - 0.89

 = 0.01 seconds.

Hope it helps.

Answer 2

Answer ⇒ Time period of physical pendulum is 0.90 seconds and difference is 0.01 seconds.

Explanation ⇒

We know that the time period of a physical pendulum is given by,

T = 2\pi \sqrt{\frac{i}{mgl} }T=2π

mgl

i

, where i is the moment of inertia about the support.

We know that the M.I. about the centre of hollow sphere = 2/3 mr² Thu, by parallel axis theorem, it can be obtained that,

Moment of inertia about the support about the support,

i = 2mr²/3 + ml²

∴ i = m(2r²/3 + l²)

∴ i = m(2(0.02)²/3 + (0.20²))

[∵ r = 0.02 cm and l = 0.18 + 0.02 = 0.20 cm.]

∴ i = 0.040m

∴ T = 2π√(0.040m/m × 9.8× 0.20)

∴ T = 0.90 s

Hence, the time period of oscillations = 0.90 seconds.