In a mass-spring system, a mass $ m $ is attached to a spring with a spring constant $ k $.
When the mass is displaced from its equilibrium position and released, it oscillates back and forth.
The period of these oscillations is given by:
$$ T = 2\pi \sqrt{\frac{m}{k}} $$
Tip
When solving for the period $T$, ensure that the spring constant $k$ is in $\text{N m}^{-1}$ and the mass $m$ is in $\text{kg}$ for consistent SI units.
Derivation of the Formula
Restoring force: The spring exerts a force proportional to the displacement $ x $, described by Hooke’s Law: $$ F = -kx $$
Newton’s Second Law: The force causes an acceleration $ a $, so $$ ma = -kx $$
Acceleration in SHM: In simple harmonic motion, $ a = -\omega^2 x $, where $ \omega $ is the angular frequency.
Equating the two expressions: $$ ma = -kx \implies a = -\frac{k}{m}x $$
Angular frequency: Comparing with $ a = -\omega^2 x $, we find $$ \omega^2 = \frac{k}{m} $$
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
The period increases with mass $ m $. A heavier mass oscillates more slowly.
The period decreases with a stiffer spring $ k $. A stiffer spring pulls the mass back more quickly.
Common Mistake
Example question
A 0.5 kg mass is attached to a spring with a spring constant of $200 \text{ N m}^{-1}$. What is the period of oscillation?
Solution
Identify the values: $ m = 0.5 \, \text{kg} $, $ k = 200 \, \text{N m}^{-1} $
Use the formula:
$$T = 2\pi \sqrt{\frac{m}{k}}$$
$$ = 2\pi \sqrt{\frac{0.5}{200}}$$
$$ \approx 0.31 \, \text{s}$$
Period of Oscillations in a Simple Pendulum
A simple pendulum consists of a mass $ m $ (the bob) attached to a string of length $ l $.
When displaced and released, it swings back and forth under the influence of gravity.
The period of a simple pendulum is given by: $$ T = 2\pi \sqrt{\frac{l}{g}} $$
Derivation of the Formula
Restoring force: When the pendulum is displaced by a small angle $ \theta $, the component of gravitational force along the arc has magnitude $ mg\sin\theta $ and acts toward the equilibrium position.
Small angle approximation: For small angles, $ \sin \theta \approx \theta $ (in radians), so: $$ F \approx -mg \theta $$
Angular displacement: The arc length $ x $ is related to the angle by $ x = l\theta $, so: $$ \theta = \frac{x}{l} $$
Substitute: $$ F \approx -mg \frac{x}{l} = -\frac{mg}{l} x $$
Newton’s Second Law: Applying $ F = ma $ along the arc gives: $$ m\frac{d^2x}{dt^2} = -\frac{mg}{l} x $$
Angular frequency: Comparing with $ a = -\omega^2 x $, we find $$ \omega^2 = \frac{g}{l} $$
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
The period depends only on the length of the pendulum and the gravitational acceleration $ g $.
It is independent of the mass of the bob and the amplitude (as long as the amplitude is small).
Example question
Calculate the period of a pendulum with a length of 1.5 m on Earth ($ g = 9.81 \, \text{m s}^{-2} $).
Solution
Identify the values: $ l = 1.5 \, \text{m} $, $ g = 9.81 \, \text{m s}^{-2} $.
Use the formula:
$$T = 2\pi \sqrt{\frac{l}{g}} $$
$$= 2\pi \sqrt{\frac{1.5}{9.81}}$$
$$ \approx 2.46 \, \text{s}$$
Note
The formula $ T = 2\pi \sqrt{\frac{l}{g}} $ is valid only for small angles (typically less than 15°).
For larger angles, the approximation $ \sin \theta \approx \theta $ breaks down, and the period increases slightly.
Comparing the Two Systems
Both the mass-spring system and the simple pendulum exhibit simple harmonic motion, but their periods depend on different factors:
Mass-Spring System: The period depends on the mass $ m $ and the spring constant $ k $.
Simple Pendulum: The period depends on the length $ l $ and gravitational acceleration $ g $.
Tip
Remember that the period of oscillation is independent of amplitude in both systems, a defining feature of simple harmonic motion.