A mass-spring system has k=200 N/m and m=0.5 kg — find period and frequency

Question

A mass-spring system has a spring constant $k = 200$ N/m and a mass $m = 0.5$ kg attached. Find: (a) the time period of oscillation, and (b) the frequency.

Solution — Step by Step

For a mass $m$ on a spring with constant $k$ , the time period of SHM is:

T = 2\pi\sqrt{\frac{m}{k}}

This formula comes from equating the restoring force $F = -kx$ with $ma$ , giving $a = -(k/m)x$ . Comparing with $a = -\omega^2 x$ (definition of SHM), we get $\omega = \sqrt{k/m}$ , and $T = 2\pi/\omega$ .

T = 2\pi\sqrt{\frac{0.5}{200}} = 2\pi\sqrt{\frac{1}{400}} = 2\pi \times \frac{1}{20} = \frac{2\pi}{20} = \frac{\pi}{10}

T = \frac{\pi}{10} \approx \frac{3.14}{10} \approx 0.314 \text{ s}

Frequency is the reciprocal of the time period:

f = \frac{1}{T} = \frac{10}{\pi} \approx \frac{10}{3.14} \approx 3.18 \text{ Hz}

Alternatively: $\omega = \sqrt{k/m} = \sqrt{200/0.5} = \sqrt{400} = 20$ rad/s, so $f = \omega/(2\pi) = 20/(2\pi) = 10/\pi \approx 3.18$ Hz.

Why This Works

The spring-mass system is the prototypical SHM oscillator. The spring provides the restoring force (proportional to displacement, directed towards equilibrium), and the mass provides inertia. A stiffer spring (larger $k$ ) oscillates faster. A heavier mass oscillates slower.

The formula $T = 2\pi\sqrt{m/k}$ shows $T \propto \sqrt{m}$ and $T \propto 1/\sqrt{k}$ . Doubling the mass increases $T$ by $\sqrt{2}$ ; quadrupling the spring constant halves $T$ .

Alternative Method — Direct Angular Frequency

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20 \text{ rad/s}

T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10} \approx 0.314 \text{ s}

f = \frac{\omega}{2\pi} = \frac{20}{2\pi} \approx 3.18 \text{ Hz}

Common Mistake

Students sometimes use $T = 2\pi\sqrt{k/m}$ instead of $T = 2\pi\sqrt{m/k}$ — they invert the fraction. Remember: heavier mass = slower oscillation = larger $T$ , so $m$ must be in the numerator. You can verify: if $m$ increases, $T$ should increase. With $m$ in the numerator, $T$ increases with $m$ . ✓

Important: the period of a spring-mass system does NOT depend on the amplitude of oscillation (as long as it stays within the elastic limit). This is a key property of SHM — isochronous means “equal time” regardless of amplitude. Pendulums also show this for small amplitudes. CBSE/JEE uses this fact in conceptual questions.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Angular Frequency

Common Mistake

Try These Next