Period of Oscillations in a Mass-Spring System
- 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:
$$
T = 2\pi \sqrt{\frac{m}{k}}
$$
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.
- Students often assume the period depends on the amplitude of oscillation.
- In SHM, the period is independent of amplitude.
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
- Identifythe 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.
