Oscillations • Topic 3 of 3

Spring & Pendulum Systems

Two systems show up again and again as textbook examples of SHM: a block attached to a spring and a simple pendulum. Both produce a restoring force proportional to displacement, so both oscillate harmonically (for small displacements in the pendulum's case).

Spring–mass system. A mass $m$ attached to a spring of force constant $k$ obeys Hooke's law, $F=-kx$. Comparing with $F=-m\omega^2 x$ gives $\omega^2=\frac{k}{m}$, so $\omega=\sqrt{\frac{k}{m}}$ and the time period is $T=2\pi\sqrt{\frac{m}{k}}$. Notice that $T$ increases with mass and decreases with stiffness, and — importantly — the period of an ideal spring–mass oscillator is independent of $g$, so it would be the same on the Moon as on Earth.

  • Springs in series: the effective force constant is smaller, $\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}$. For two identical springs of constant $k$, $k_s=\frac{k}{2}$ — the combination is softer, so the period is longer.
  • Springs in parallel: the effective force constant adds, $k_p=k_1+k_2$. For two identical springs, $k_p=2k$ — the combination is stiffer, so the period is shorter.

Simple pendulum. A point mass (bob) on a light inextensible string of length $L$, displaced by a small angle, experiences a restoring force $-mg\sin\theta\approx -mg\theta$ (using the small-angle approximation $\sin\theta\approx\theta$ for $\theta$ in radians, valid for angles up to roughly $10^\circ$). This leads to SHM with $\omega=\sqrt{\frac{g}{L}}$ and time period $T=2\pi\sqrt{\frac{L}{g}}$. The period depends only on the length and on $g$ — it is independent of the mass of the bob and of the amplitude (for small swings). A longer pendulum, or a place with smaller $g$, swings more slowly.

Damped oscillations. Real oscillators lose energy to friction and air resistance. A damping force proportional to velocity, $F_d=-bv$, causes the amplitude to decay exponentially with time, $A(t)=A_0 e^{-bt/2m}$, while the frequency shifts slightly below the natural value. Light damping gives a slow ring-down; heavy damping (over-damping) returns the system to rest without oscillating.

Forced oscillations and resonance. If an external periodic force of frequency $\omega_d$ drives the oscillator, the system settles into steady oscillation at the driving frequency. When the driving frequency matches the system's natural frequency ($\omega_d\approx\omega_0$), the amplitude grows very large — this is resonance. Resonance explains why a pushed swing builds up, why soldiers break step on bridges, and how radios tune in. The less damped the system, the taller and narrower the resonance peak.

Spring-mass system and simple pendulummT = 2π√(m/k)LθT = 2π√(L/g)
Solved Examples
1
Worked Example
A 0.5 kg mass attached to a spring of force constant $200\ \text{N/m}$ oscillates on a frictionless surface. Find the time period of oscillation.
Solution
  1. Step 1: Use $T=2\pi\sqrt{\frac{m}{k}}$.
  2. Step 2: Substitute: $T=2\pi\sqrt{\frac{0.5}{200}}=2\pi\sqrt{0.0025}$.
  3. Step 3: $\sqrt{0.0025}=0.05$, so $T=2\pi\times0.05=0.1\pi\approx0.314\ \text{s}$.

Answer: $T\approx0.314\ \text{s}$.

2
Worked Example
Find the length of a simple pendulum whose time period is 2 s (a 'seconds pendulum'). Take $g=9.8\ \text{m/s}^2$.
Solution
  1. Step 1: From $T=2\pi\sqrt{\frac{L}{g}}$, square both sides: $T^2=4\pi^2\frac{L}{g}$, so $L=\frac{gT^2}{4\pi^2}$.
  2. Step 2: Substitute: $L=\frac{9.8\times2^2}{4\pi^2}=\frac{9.8\times4}{39.48}=\frac{39.2}{39.48}$.
  3. Step 3: $L\approx0.993\ \text{m}\approx0.99\ \text{m}$.

Answer: $L\approx0.99\ \text{m}$ (about 1 m).

3
Worked Example
Two springs of force constants $300\ \text{N/m}$ and $600\ \text{N/m}$ are connected in series. Find the effective force constant of the combination.
Solution
  1. Step 1: For springs in series, $\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}$.
  2. Step 2: $\frac{1}{k_s}=\frac{1}{300}+\frac{1}{600}=\frac{2+1}{600}=\frac{3}{600}=\frac{1}{200}$.
  3. Step 3: $k_s=200\ \text{N/m}$ (softer than either spring).

Answer: $k_s=200\ \text{N/m}$.

4
Worked Example
A simple pendulum has a period of 2 s on Earth. What will its period be on the Moon, where $g$ is one-sixth of its value on Earth?
Solution
  1. Step 1: $T\propto\frac{1}{\sqrt{g}}$ for fixed length, so $\frac{T_{moon}}{T_{earth}}=\sqrt{\frac{g_{earth}}{g_{moon}}}=\sqrt{6}$.
  2. Step 2: $T_{moon}=T_{earth}\sqrt{6}=2\times\sqrt{6}$.
  3. Step 3: $\sqrt{6}=2.449$, so $T_{moon}=2\times2.449\approx4.9\ \text{s}$.

Answer: $T_{moon}\approx4.9\ \text{s}$.

5
Worked Example
Two identical springs each of force constant $k$ are joined in parallel and a mass $m$ is hung from the combination. Express the time period of vertical oscillation.
Solution
  1. Step 1: For two identical springs in parallel, the effective constant is $k_p=k+k=2k$.
  2. Step 2: Substitute into $T=2\pi\sqrt{\frac{m}{k_p}}$.
  3. Step 3: $T=2\pi\sqrt{\frac{m}{2k}}$ — shorter than for a single spring because the combination is stiffer.

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

6
Worked Example
The amplitude of a damped oscillator falls to half its initial value in 20 s. Using $A=A_0 e^{-bt/2m}$, find the time at which the amplitude becomes one-quarter of the initial value.
Solution
  1. Step 1: The amplitude decays exponentially, so equal factors of $\frac{1}{2}$ take equal times.
  2. Step 2: Falling to $\frac{1}{2}$ takes 20 s; falling from $\frac{1}{2}$ to $\frac{1}{4}$ takes another 20 s.
  3. Step 3: Total time for one-quarter $=20+20=40\ \text{s}$.

Answer: $t=40\ \text{s}$.

Key Points

  • Spring–mass system: $T=2\pi\sqrt{\frac{m}{k}}$ — increases with mass, decreases with stiffness, and is independent of $g$.
  • Springs in series give a smaller constant ($\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}$); in parallel they add ($k_p=k_1+k_2$).
  • Simple pendulum (small angle): $T=2\pi\sqrt{\frac{L}{g}}$ — independent of the mass and amplitude of the bob.
  • Damped oscillations: a velocity-dependent force makes the amplitude decay as $A(t)=A_0 e^{-bt/2m}$.
  • Resonance occurs when the driving frequency equals the natural frequency, producing maximum amplitude.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The time period of a spring–mass system is given by:
Explanation: Comparing $F=-kx$ with $F=-m\omega^2 x$ gives $\omega=\sqrt{k/m}$, so $T=2\pi\sqrt{m/k}$.
Q2.The time period of a simple pendulum is independent of the:
Explanation: $T=2\pi\sqrt{L/g}$ contains only $L$ and $g$; the bob\u2019s mass does not appear, so the period is independent of mass.
Q3.When two identical springs of constant $k$ are connected in parallel, the effective force constant is:
Explanation: Springs in parallel add their constants, so $k_p=k+k=2k$.
Q4.The condition for resonance in a forced oscillation is that the driving frequency is:
Explanation: Amplitude is maximum when the driving frequency matches the natural frequency $\omega_0$ — this is resonance.
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