C.1.2 Time period of oscillatory systems Notes

Period of Oscillations in a Mass-Spring System

1. In a mass-spring system, a mass $ m $ is attached to a spring with a spring constant $ k $.
2. When the mass is displaced from its equilibrium position and released, it oscillates back and forth.
3. The period of these oscillations is given by:

$$
T = 2\pi \sqrt{\frac{m}{k}}
$$

Derivation of the Formula

1. Restoring force: The spring exerts a force proportional to the displacement $ x $, described by Hooke’s Law: $$ F = -kx $$
2. Newton’s Second Law: The force causes an acceleration $ a $, so $$ ma = -kx $$
3. Acceleration in SHM: In simple harmonic motion, $ a = -\omega^2 x $, where $ \omega $ is the angular frequency.
4. Equating the two expressions: $$
   ma = -kx \implies a = -\frac{k}{m}x
   $$
5. Angular frequency: Comparing with $ a = -\omega^2 x $, we find $$ \omega^2 = \frac{k}{m} $$
6. Period: The period $ T $ is related to angular frequency by $ T = \frac{2\pi}{\omega} $, so: $$
   T = 2\pi \sqrt{\frac{m}{k}}
   $$

Key Insights

1. The period increases with mass $ m $. A heavier mass oscillates more slowly.
2. The period decreases with a stiffer spring $ k $. A stiffer spring pulls the mass back more quickly.

Period of Oscillations in a Simple Pendulum

1. A simple pendulum consists of a mass $ m $ (the bob) attached to a string of length $ l $.
2. When displaced and released, it swings back and forth under the influence of gravity.
3. The period of a simple pendulum is given by: $$
   T = 2\pi \sqrt{\frac{l}{g}}
   $$

Derivation of the Formula

1. Restoring force: When the pendulum is displaced by a small angle $ \theta $, the component of gravitational force acting along the arc is: $$ F = -mg \sin \theta $$
2. Small angle approximation: For small angles, $ \sin \theta \approx \theta $ (in radians), so: $$ F \approx -mg \theta $$
3. Angular displacement: The arc length $ x $ is related to the angle by $ x = l\theta $, so: $$ \theta = \frac{x}{l} $$
4. Substitute: $$ F \approx -mg \frac{x}{l} = -\frac{mg}{l} x $$
5. Newton’s Second Law: For rotational motion, $ F = ma $ becomes: $$ m\frac{d^2x}{dt^2} = -\frac{mg}{l} x $$
6. Angular frequency: Comparing with $ a = -\omega^2 x $, we find $$ \omega^2 = \frac{g}{l} $$
7. Period: The period $ T $ is related to angular frequency by $ T = \frac{2\pi}{\omega} $, so: $$
   T = 2\pi \sqrt{\frac{l}{g}}
   $$

Key Insights

1. The period depends only on the length of the pendulum and the gravitational acceleration$ g $.
2. It is independent of the mass of the bob and the amplitude (as long as the amplitude is small).

Comparing the Two Systems

Both the mass-spring system and the simple pendulum exhibit simple harmonic motion, but their periods depend on different factors:

1. Mass-Spring System: The period depends on the mass $ m $ and the spring constant $ k $.
2. Simple Pendulum: The period depends on the length $ l $ and gravitational acceleration $ g $.