Stretching Springs, SHM & Pendulums

Hooke's law, elastic energy, simple harmonic motion, mathematical and elastic pendulums

1. Hooke's law

When a spring is stretched or compressed, the force it exerts is proportional to the extension (or compression) \(x\) from its natural length, provided the spring is not permanently deformed:

\[ F = -k x \]

Here \(k\) is the spring constant (unit: N/m). The minus sign means the force opposes the displacement (restoring force).

2. Elastic potential energy

Work done to stretch the spring from \(0\) to \(x\): \(W = \int_0^x F'\,dx' = \int_0^x kx'\,dx' = \frac{1}{2}kx^2\). This is stored as elastic potential energy:

\[ U = \frac{1}{2} k x^2 \]

Proof: Force to extend spring by \(dx'\) is \(kx'\). Work \(dW = kx'\,dx'\). Total work \(W = \int_0^x kx'\,dx' = k\left[\frac{x'^2}{2}\right]_0^x = \frac{1}{2}kx^2\). By definition this equals the potential energy stored.

3. Simple harmonic motion (SHM) of a mass on a spring

For a mass \(m\) on a frictionless horizontal surface attached to a spring of constant \(k\), Newton's second law gives \(m\ddot{x} = -kx\), so \(\ddot{x} + \frac{k}{m}x = 0\). This is the SHM equation with angular frequency \(\omega = \sqrt{k/m}\).

\[ \omega = \sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}}, \qquad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \]

Proof: Let \(x = A\cos(\omega t + \phi)\). Then \(\ddot{x} = -\omega^2 A\cos(\omega t + \phi) = -\omega^2 x\). Substituting into \(m\ddot{x} = -kx\): \(-m\omega^2 x = -kx\), so \(\omega^2 = k/m\), hence \(\omega = \sqrt{k/m}\) and \(T = 2\pi/\omega = 2\pi\sqrt{m/k}\).

4. Mathematical (simple) pendulum

A point mass \(m\) on a massless string of length \(L\) in a gravitational field. For small angles \(\theta\), the restoring torque is \(mgL\sin\theta \approx mgL\theta\), giving SHM with angular frequency \(\omega = \sqrt{g/L}\).

\[ T = 2\pi\sqrt{\frac{L}{g}}, \qquad \omega = \sqrt{\frac{g}{L}}, \qquad f = \frac{1}{2\pi}\sqrt{\frac{g}{L}} \]

Proof (small angle): Tangential force \(F_t = -mg\sin\theta \approx -mg\theta\). Displacement along arc \(s = L\theta\), so \(a = \ddot{s} = L\ddot{\theta}\). Newton: \(mL\ddot{\theta} = -mg\theta\), so \(\ddot{\theta} + (g/L)\theta = 0\). Hence \(\omega^2 = g/L\), \(T = 2\pi/\omega = 2\pi\sqrt{L/g}\).

5. Elastic pendulum (spring pendulum)

A mass attached to a spring (instead of a rigid string) that can stretch. Motion is a combination of: (1) oscillation of the spring length (vertical SHM if hung vertically) and (2) swinging (pendulum-like). For small oscillations about the vertical, the period of vertical motion is \(T_s = 2\pi\sqrt{m/k}\) (with \(k\) the spring constant); the period of swing depends on the effective length. If the spring is horizontal and mass slides on frictionless surface, it is pure SHM with \(T = 2\pi\sqrt{m/k}\).

Vertical spring (mass hanging): equilibrium extension \(x_0 = mg/k\). Oscillation about equilibrium: same \(T = 2\pi\sqrt{m/k}\). Elastic pendulum (spring + gravity): two modes; approximate formulas involve both \(k\) and \(L\).

Formula summary

Hooke: \(F = -kx\)
Elastic PE: \(U = \frac{1}{2}kx^2\)
Spring SHM: \(T = 2\pi\sqrt{m/k}\), \(\omega = \sqrt{k/m}\)
Simple pendulum: \(T = 2\pi\sqrt{L/g}\) (small angles)

Worked examples

Simple — Spring force

A spring has \(k = 100\,\text{N/m}\). How much force is needed to stretch it by \(0.05\,\text{m}\)?

Solution: \(F = kx = 100 \times 0.05 = 5\,\text{N}\). (We use magnitude; the applied force opposes the restoring force.)

Simple — Elastic PE

The same spring is compressed by \(0.02\,\text{m}\). Find the stored elastic potential energy.

Solution: \(U = \frac{1}{2}kx^2 = \frac{1}{2}(100)(0.02)^2 = \frac{1}{2}(100)(0.0004) = 0.02\,\text{J}\).

Intermediate — Period of spring

A mass of \(0.5\,\text{kg}\) is attached to a spring of constant \(200\,\text{N/m}\). Find the period of oscillation.

Solution: \(T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/200} = 2\pi\sqrt{0.0025} = 2\pi(0.05) = 0.314\,\text{s}\).

Intermediate — Simple pendulum

A simple pendulum has length \(1.6\,\text{m}\). Find its period. Use \(g = 10\,\text{m/s}^2\).

Solution: \(T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1.6/10} = 2\pi\sqrt{0.16} = 2\pi(0.4) = 2.51\,\text{s}\).

Advanced — Spring and energy

A block of mass \(2\,\text{kg}\) is attached to a spring (\(k = 50\,\text{N/m}\)) on a frictionless surface. The block is pulled \(0.2\,\text{m}\) from equilibrium and released. Find (a) the maximum speed and (b) the speed when displacement is \(0.1\,\text{m}\).

Solution: (a) At release, total energy \(E = \frac{1}{2}kA^2 = \frac{1}{2}(50)(0.2)^2 = 1\,\text{J}\). At equilibrium, all energy is kinetic: \(\frac{1}{2}mv_{max}^2 = 1\), so \(v_{max} = \sqrt{2/2} = 1\,\text{m/s}\). (b) At \(x = 0.1\,\text{m}\): \(\frac{1}{2}kx^2 + \frac{1}{2}mv^2 = 1\). So \(\frac{1}{2}(50)(0.01) + \frac{1}{2}(2)v^2 = 1\), \(0.25 + v^2 = 1\), \(v^2 = 0.75\), \(v = 0.866\,\text{m/s}\).

Advanced — Pendulum length from period

A simple pendulum has a period of \(2\,\text{s}\). Find its length. Use \(g = 9.8\,\text{m/s}^2\).

Solution: \(T = 2\pi\sqrt{L/g}\) ⇒ \(T^2 = 4\pi^2 L/g\) ⇒ \(L = \frac{gT^2}{4\pi^2} = \frac{9.8 \times 4}{4\pi^2} = \frac{39.2}{39.48} \approx 0.993\,\text{m}\).

Problems

Simple, Intermediate, and Advanced. Click Show solution to reveal.