For a mass m attached to a horizontal spring with spring constant k undergoing simple harmonic motion, how is the angular frequency ω related to m and k?

For a mass m attached to a horizontal spring with spring constant k undergoing simple harmonic motion, how is the angular frequency ω related to m and k?

• A. ω = sqrt(m/k)
• B. ω = sqrt(k/m)
• C. ω = k/m
• D. ω = sqrt(m)/sqrt(k)

Answer: (B)

The angular frequency of a mass on a horizontal spring in simple harmonic motion comes from balancing restoring force and inertia. The restoring force is F = -kx, so Newton’s second law gives m d^2x/dt^2 = -k x. This rearranges to d^2x/dt^2 + (k/m) x = 0, which matches the standard SHM form x'' + ω^2 x = 0. Therefore ω^2 = k/m, so ω = sqrt(k/m). This means stiffer springs (larger k) make the system vibrate faster, while more massive objects (larger m) slow the motion down. You can also see the period, T = 2π/ω = 2π sqrt(m/k), which aligns with that intuition.

The angular frequency of a mass on a horizontal spring in simple harmonic motion comes from balancing restoring force and inertia. The restoring force is F = -kx, so Newton’s second law gives m d^2x/dt^2 = -k x. This rearranges to d^2x/dt^2 + (k/m) x = 0, which matches the standard SHM form x'' + ω^2 x = 0. Therefore ω^2 = k/m, so ω = sqrt(k/m).

This means stiffer springs (larger k) make the system vibrate faster, while more massive objects (larger m) slow the motion down. You can also see the period, T = 2π/ω = 2π sqrt(m/k), which aligns with that intuition.