Mass-Spring Systems and the Simple Pendulum

A mass m on a spring of constant k oscillates with period T = 2π√(m ÷ k), and a simple pendulum of length L swinging through small angles has period T = 2π√(L ÷ g); neither period depends on the amplitude (for small amplitude). In brief: a stiffer spring or smaller mass gives a shorter period; a longer pendulum or weaker gravity gives a longer period. The period of a pendulum does not depend on the mass of the bob. A vertical spring oscillates about the new equilibrium position where the weight is balanced by the spring force, with the same period as a horizontal one.

The two standard examples of simple harmonic motion are used in clocks, in measuring g, in suspension systems and in vibrations of atoms. This guide derives the period of a mass on a spring and a simple pendulum, explains the conditions for the formulas, compares the two systems in a table, works through numerical examples including measuring g and corrects common mistakes.

How does a mass on a spring oscillate?

For a mass m on a horizontal frictionless surface attached to a spring, the restoring force is F = −kx, so a = −(k ÷ m)x, which is SHM with ω = √(k ÷ m). The period is

T = 2π √(m ÷ k), and the frequency f = (1 ÷ 2π) √(k ÷ m).

Quadrupling the mass doubles the period. For a vertical spring, gravity shifts the equilibrium by Δ = mg ÷ k (the static extension), but the motion about this new equilibrium is SHM with the same ω, since the constant force of gravity does not change the form of the equation. The period can also be written as T = 2π√(Δ ÷ g), so the static extension gives the period directly. A spring with a mass of its own contributes about one-third of its mass to the effective mass. For springs in parallel, k = k1 + k2; in series, 1 ÷ k = 1 ÷ k1 + 1 ÷ k2.

How does a simple pendulum swing?

A simple pendulum consists of a small mass (the bob) on a light, inextensible string of length L. When it is displaced by an angle θ, the component of weight along the arc, −mg sin θ, restores it. For small angles (less than about 10°), sin θ ≈ θ (in radians), and the arc displacement is x = Lθ, so the restoring force is approximately −(mg ÷ L)x. This is SHM with keff = mg ÷ L, so ω = √(g ÷ L) and

T = 2π √(L ÷ g).

The mass cancels. The period depends only on the length and g. A pendulum 1.00 m long has a period of about 2.0 s on the Earth (the “seconds pendulum” has a half-period of 1 s and a length of 0.994 m). For larger amplitudes the period increases slightly: for a swing of 10° amplitude, by only 0.2 percent; for 30°, by about 1.7 percent. The length is measured from the point of suspension to the center of the bob. Because T depends on g, a pendulum can be used to measure g: g = 4π2L ÷ T2, and a plot of T2 against L is a straight line with slope 4π2 ÷ g.

Period squared against length for a simple pendulum is a straight line 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0 1 2 3 4 5 Pendulum length L (m) Period squared T2 (s2) 0.800 m, 3.24 s² slope = 4π2 / g
For a simple pendulum, T² = (4π² ÷ g)L is a straight line through the origin whose slope gives g. The line is drawn for g = 9.81 m/s², a slope of 4.02 s²/m, so a 1.00 m pendulum has T² = 4.02 s² (T = 2.01 s). The dot is the measurement from the worked example, L = 0.800 m and T = 1.80 s, which gives g = 9.75 m/s².

The two systems compared

Mass-spring system and simple pendulum
Feature Mass on a spring Simple pendulum
Restoring force−kx (exact, within the elastic limit)−mg sin θ ≈ −(mg ÷ L)x (small angles)
Angular frequency ω√(k ÷ m)√(g ÷ L)
Period T2π√(m ÷ k)2π√(L ÷ g)
Depends on mass?YesNo
Depends on g?No (horizontal or vertical)Yes

Worked example: periods and measuring g

  1. Spring. A 0.50 kg mass on a spring with k = 200 N/m: T = 2π√(0.50 ÷ 200) = 2π × 0.050 = 0.314 s, f = 3.18 Hz. With a 2.0 kg mass: T = 2π√(2.0 ÷ 200) = 0.628 s (double).
  2. Vertical spring. The 0.50 kg mass hangs at rest: static extension = mg ÷ k = 0.50 × 9.8 ÷ 200 = 0.0245 m. The period is the same: T = 2π√(0.0245 ÷ 9.8) = 0.314 s.
  3. Pendulum. L = 1.00 m: T = 2π√(1.00 ÷ 9.81) = 2.01 s. For T = 2.00 s: L = gT2 ÷ 4π2 = 9.81 × 4.00 ÷ 39.5 = 0.994 m.
  4. On the Moon. g = 1.62 m/s2: T = 2π√(1.00 ÷ 1.62) = 4.94 s, which is 2.46 times longer.
  5. Measuring g. A pendulum with L = 0.800 m takes 36.0 s for 20 swings (T = 1.80 s). g = 4π2L ÷ T2 = 39.48 × 0.800 ÷ 3.24 = 9.75 m/s2. Timing 20 swings and not one reduces the effect of the reaction time (±0.2 s becomes ±0.01 s in T).
  6. Maximum speed of a pendulum. For a 1.00 m pendulum with an amplitude of 5.0 cm: vmax = Aω = 0.050 × √(9.81 ÷ 1.00) = 0.050 × 3.13 = 0.157 m/s.
0.50 kg kΔ = 4.9 N mg = 4.9 N

How were they used in timekeeping?

Galileo studied pendulums in the 1580s to 1600s. Christiaan Huygens built the first pendulum clock in 1656 and published its theory in 1673; it improved accuracy from about 15 minutes per day to 15 seconds per day. In 1851, Léon Foucault hung a 67 m pendulum in the Panthéon in Paris to show the Earth’s rotation. Pendulums were used to survey gravity and discover that g varies with latitude; mass-spring systems are the basis of suspension in cars and accelerometers.

Which mistakes are common?

Mistake 1: Thinking that a heavier pendulum bob swings more slowly

The period does not depend on the mass of the bob.

Mistake 2: Using the formula for large amplitudes

The simple formula is accurate only for small angles (below about 10°).

Mistake 3: Using the string length instead of the length to the center of the bob

L is measured to the center of mass of the bob.

Mistake 4: Thinking that a vertical spring has a different period

Gravity shifts the equilibrium but does not change the period.