Mass-spring system, simple pendulum and energy in SHM

Physics · Oscillations. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A mass on a spring and a simple pendulum are the two standard examples of SHM. For the mass-spring system the restoring force is F = −kx, so the period depends only on the mass and the spring constant. For the simple pendulum swinging through a small angle, the period depends only on its length and on g. The mass of the bob and the amplitude make no difference.

During the motion, energy changes back and forth between kinetic and potential. At the extreme positions it is all potential energy. At the mean position it is all kinetic energy. With no friction the total energy stays constant and is proportional to the square of the amplitude.

Rules to remember

mass-spring system: T = 2π√(m/k), where k is the spring constant in N m⁻¹; it does not depend on g
simple pendulum: T = 2π√(l/g); it does not depend on the mass of the bob
a second's pendulum has a period of 2.0 s (length about 0.99 m where g = 9.8 m s⁻²)
potential energy at displacement x: P.E. = ½kx²
kinetic energy at displacement x: K.E. = ½k(x₀² − x²)
total energy = ½kx₀², constant at every point
K.E. = P.E. when x = x₀/√2

Common mistake

The period depends on a square root. Making the mass or the length four times as large only doubles the period. Also, taking a mass-spring system to the Moon does not change its period, but a pendulum there swings more slowly.

Worked example

A mass on a spring of spring constant 80 N m⁻¹ oscillates with amplitude 0.050 m. Find its kinetic energy when the displacement is 0.030 m.

Total energy = ½kx₀² = ½ × 80 × 0.050² = 0.100 J.
P.E. at x = 0.030 m: ½kx² = ½ × 80 × 0.030² = 0.036 J.
K.E. = total − P.E. = 0.100 − 0.036 = 0.064 J.

Answer: 0.064 J

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Chapter 7

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