Question

Obtain an expression for period of simple pendulum. On which factors it depends?

Answer 1

Let m is Mass of the bob.
L is Length of mass-less string.
Free-body diagram to the forces acting on the bob,
θ − angle made by the string with the vertical.
T is tension along the string
g is acceleration due to gravity

Radial acceleration = ω²L
Net radial force = T − mg cosθ
Tangential acceleration is provided by mg sinθ.
Torques, τ = − L (mg sinθ)
According to Newton’s law of rotational motion,
τ = Iα
Where, I is the moment of inertia
α − Angular acceleration
∴ Iα = − mg sinθ L
If θ is very small, then
sinθ ≈ θ
so, Iα = − mgθL = -(mgL)θ
We know, it will follow simple harmonic motion when $\alpha=-\omega^2\theta$
So, $I(-\omega^2\theta)=-(mgL)\theta$

Moment of inertia , I = mL²

So, $mL^2\omega^2=mgL$
$\omega=\sqrt{\frac{g}{L}}$

$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{L}{g}}$

Here it is clear that time period of simple pendulum is directly proportional to square root of length of pendulum and inversely proportional to square root of acceleration due to gravity.

Answer 2

Answer:

θ − angle made by the string with the vertical.

T is tension along the string

g is acceleration due to gravity

Radial acceleration = ω²L

Net radial force = T − mg cosθ

Tangential acceleration is provided by mg sinθ.

Torques, τ = − L (mg sinθ)

According to Newton’s law of rotational motion,

τ = Iα

Where, I is the moment of inertia

α − Angular acceleration

∴ Iα = − mg sinθ L

If θ is very small, then

sinθ ≈ θ

so, Iα = − mgθL = -(mgL)θ

We know, it will follow simple harmonic motion when

\alpha=-\omega^2\theta

So,

I(-\omega^2\theta)=-(mgL)\theta

Moment of inertia , I = mL²

So,

mL^2\omega^2=mgL

\omega=\sqrt{\frac{g}{L}}

T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{L}{g}}

ANSWER

Explanation:

$.$