For a mass m attached to a horizontal spring with constant k displaced by x and released, what is the angular frequency ω of the resulting simple harmonic motion?

For a mass m attached to a horizontal spring with constant k displaced by x and released, what is the angular frequency ω of the resulting simple harmonic motion?

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

Answer: (B)

When a mass on a horizontal spring experiences a restoring force proportional to displacement, the motion is simple harmonic. The force is F = -k x, and applying Newton’s second law gives m d^2x/dt^2 = -k x. This leads to the standard form d^2x/dt^2 + (k/m) x = 0, whose solution oscillates with angular frequency ω where ω^2 = k/m. Therefore, ω = sqrt(k/m). This shows how stiffness k and mass m set the rate of oscillation: bigger k makes the system stiffer and increases ω, while bigger m makes it heavier and decreases ω. The period is T = 2π sqrt(m/k). The other expressions don’t fit because they either invert the ratio or yield the wrong dimensionality for angular frequency. Only sqrt(k/m) has the correct dependence and units.

When a mass on a horizontal spring experiences a restoring force proportional to displacement, the motion is simple harmonic. The force is F = -k x, and applying Newton’s second law gives m d^2x/dt^2 = -k x. This leads to the standard form d^2x/dt^2 + (k/m) x = 0, whose solution oscillates with angular frequency ω where ω^2 = k/m. Therefore, ω = sqrt(k/m).

This shows how stiffness k and mass m set the rate of oscillation: bigger k makes the system stiffer and increases ω, while bigger m makes it heavier and decreases ω. The period is T = 2π sqrt(m/k).

The other expressions don’t fit because they either invert the ratio or yield the wrong dimensionality for angular frequency. Only sqrt(k/m) has the correct dependence and units.