Simple harmonic motion practice questions.

Which condition must be satisfied for the motion of a particle to be simple harmonic?

• Its acceleration is constant in magnitude and direction.
• Its acceleration is proportional to its velocity and opposes the motion.
• Its acceleration is proportional to its displacement and directed toward the equilibrium position.
• Its acceleration is proportional to its displacement and directed away from the equilibrium position.

Answer: Its acceleration is proportional to its displacement and directed toward the equilibrium position.

SHM is defined by a = −ω²x: the acceleration is proportional to displacement and always points back toward equilibrium (the minus sign).

A 0.25 kg mass hangs from a spring of spring constant 100 N m⁻¹. The mass is displaced vertically and released. Calculate the period of the resulting oscillation.

• 0.050 s
• 0.31 s
• 0.63 s
• 3.2 s

Answer: 0.31 s

T = 2π√(m/k) = 2π√(0.25/100) = 2π × 0.050 = 0.31 s. Forgetting the 2π gives 0.050 s, 3.2 s is the frequency 1/T rather than the period, and 0.63 s is double the correct value.

The length of a simple pendulum is increased to four times its original value. The period of small oscillations will

• double.
• quadruple.
• halve.
• stay the same.

Answer: double.

T = 2π√(l/g), so T ∝ √l. Multiplying l by 4 multiplies the period by √4 = 2.

A particle performs SHM. At the instant it reaches maximum displacement, which statement is correct?

• Its speed is maximum and its acceleration is momentarily zero at this point.
• Its speed and acceleration are both maximum.
• Its speed and acceleration are both zero.
• Its speed is zero and the magnitude of its acceleration is maximum.

Answer: Its speed is zero and the magnitude of its acceleration is maximum.

At x = x₀ the particle momentarily stops (v = 0), while a = −ω²x has its greatest magnitude because displacement is greatest.

In simple harmonic motion, what is the phase difference between the acceleration and the displacement of the oscillating particle?

• 0 rad
• π/2 rad
• π rad
• 2π rad

Answer: π rad

Since a = −ω²x, acceleration is always opposite in sign to displacement, i.e. exactly half a cycle (π rad) out of phase.

An object performs simple harmonic motion with angular frequency ω = 4.0 rad s⁻¹. Determine the magnitude of its acceleration at the instant its displacement from equilibrium is 0.050 m.

• 0.20 m s⁻²
• 0.80 m s⁻²
• 1.6 m s⁻²
• 3.2 m s⁻²

Answer: 0.80 m s⁻²

a = ω²x = (4.0)² × 0.050 = 16 × 0.050 = 0.80 m s⁻². Using ω instead of ω² gives the common wrong answer 0.20 m s⁻².