Question

Deduce relation for time period of simple pendulum

Answer 1

Answer:

If the body displaced through a small angle (θ) and released from this position, a torque is exerted by the weight of the body to restore to its equilibrium.

τ = -mg × (d sinθ)

τ = I α

τ α = – mgdsinθ

I = d^2θ/dt^2 = – mgdsinθ

Where I = moment of inertia of a body about the axis of rotation.

d2θ/dt^2 = (mgd/I) θ [Since, sinθ ≈ θ]

ω0 = √[mgd/I].

T = 2π/ω0 = 2π × √[I/mgd]

For ‘I’, applying parallel axis theorem,

I = Icm + md2

Therefore, the time period of a physical pendulum is given by,

T = 2π × √[(Icm + md2)/mgd]

Answer 2

Answer:

T = 2π$\sqrt{\frac{l}{g} }$

Explanation:

The time period of the simple pendulum may depend on the following:

• mass (m) of the bob
• length (l) of the thread
• acceleration due to gravity (g)

Time period (T) ∝ mass of the bob × length (l) of the thread ×  acceleration due to gravity (g)

⇒ T = k×m×l×g  (where k is a dimensionless constant)

⇒ T = k($m^{a}$)($l^{b}$)($g^{c}$)

⇒ [T] = $[M]^{a}$  $[L]^{b}$  $[LT^{-2}]^{c}$      ...(i)

⇒  $[M^{0}$ $L^{0}$  $T^{1}]$ = $[M]^{a}$ $[L]^{b}$ $[LT^{-2}]^{c}$

comparing the powers

a = 0

-2c = -1

⇒ c = $\frac{1}{2}$

b + c = 0

⇒ b - $\frac{1}{2}$ = 0

⇒ b = $\frac{1}{2}$

from (i)

T = k $[m]^{0}$ $[l]^{\frac{1}{2} }$  $[g]^{\frac{-1}{2}}$

T = k$\sqrt{\frac{l}{g} }$

From experiments we know that k = 2π

∴ T = 2π$\sqrt{\frac{l}{g} }$