Name: Simple Harmonic Motion: Hooke's Law
Uploaded: 2019-07-19
Duration: 17 min 48 s
Description: View this video to learn simple harmonic motion and its associated laws using the oscillation of a spring. Who doesn't love a good oscillation now and again?

Question

Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.

Accepted Answer

Consider a simple pendulum of length $L$ suspended from a rigid support $O$ . When displaced from its initial position of rest through a small angle $θ$ in a vertical plane and released, it performs oscillations between two extremes, $B$ and $C$ , as shown in below figure. At $B$ , the forces on the bob are its weight $m \vec{g}$ and the tension $\vec{F_{1}}$ in the string.

Thus, weight $m g$ is resolved into two components:

The component $m g \cos θ$ along the string, which is balanced by the tension $T'$ and

The component $m g \sin θ$ perpendicular to the string is the restoring force acting on mass $m$ tending to return it to the equilibrium position.

Restoring force, $F = - m g \sin θ - - - - - - - (1)$
As $θ$ is very small $(< 10 °)$ , we can write
$\sin θ = θ ° ∴ F ≅ - m g θ$

From the Fig, the small angle $θ = \frac{a r c}{r a d i u s} = \frac{A B}{O B} = \frac{x}{L}$
$∴ F = - m g \frac{x}{L}$

As $m, g$ and $L$ are constant, $F \propto - x$

Thus, the net force on the bob is in the direction opposite to that of displacement $x$ of the bob from its mean position as indicated by the minus sign, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear S.H.M.
Acceleration, $a = \frac{F}{m} = - \frac{g}{L} x$
Therefore, acceleration per unit displacement
$= | \frac{a}{x} | = \frac{g}{L}$
Period of SHM,

$T = \sqrt{\frac{2 π}{a c c e l e r a t i o n p e r u n i t d i s p l a c e m e n t}} T = \frac{2 π}{\sqrt{\frac{g}{L}}} ∴ t = 2 π \sqrt{\frac{L}{g}}$
This gives the expression for the period of a simple pendulum.

where $L$ is the length of the simple pendulum and $g$ is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :

The period of a simple pendulum is directly proportional to the square root of its length.

The period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity.

The period of a simple pendulum does not depend on its mass.

The period of a simple pendulum does not depend on its amplitude (for small amplitude).
These conclusions are called the 'laws of simple pendulum'.

Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.

Important Questions on Oscillations

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