Simple Harmonic Motion: Mass on a Spring

A mass on a spring is the canonical example of simple harmonic motion (SHM). Hooke's law gives a restoring force proportional to displacement, and Newton's Second Law turns that into the harmonic equation:

$m\,\ddot x = -k\,x$

The solutions are pure sinusoids with angular frequency:

$\omega = \sqrt{\dfrac{k}{m}}$

This $\omega$ depends only on the mass and spring constant — not on the amplitude. That's the beautiful and surprising property of SHM: a small swing and a large swing complete in the same time.

The Position, Velocity, Acceleration

If you release the mass from rest at $x = A$, the motion is:

$x(t) = A\cos(\omega t)$

$v(t) = -A\omega\sin(\omega t)$

$a(t) = -A\omega^{2}\cos(\omega t) = -\omega^{2}\,x(t)$

The position and acceleration are 180° out of phase: when $x$ is at its maximum, $a$ is at its minimum (most negative). The velocity is 90° out of phase with both — maximum when $x$ is zero, zero when $x$ is at its extremes.

Step-by-Step Solution

Given: $m = 2\;\text{kg}$, $k = 50\;\text{N/m}$, amplitude $A = 0.2\;\text{m}$, released from rest at $x = A$.

Find: $\omega$, $T$, $f$, and the maximum velocity and acceleration.

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Step 1 — Compute the angular frequency.

$\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{50}{2}} = \sqrt{25} = 5\;\text{rad/s}$

Step 2 — Compute the period and frequency.

$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{5} \approx 1.257\;\text{s}$

$f = \dfrac{1}{T} = \dfrac{1}{1.257} \approx 0.796\;\text{Hz}$

So the mass completes about 0.796 cycles per second — or one cycle every 1.26 seconds.

Step 3 — Compute the maximum velocity.

The maximum speed occurs at the center ($x = 0$), where all the energy is kinetic:

$v_{\max} = A\omega = (0.2)(5) = 1.000\;\text{m/s}$

Step 4 — Compute the maximum acceleration.

The maximum acceleration occurs at the extremes ($x = \pm A$), where the spring is most stretched/compressed:

$a_{\max} = A\omega^{2} = (0.2)(25) = 5.000\;\text{m/s}^{2}$

Step 5 — Verify with energy conservation.

The total energy is $\tfrac{1}{2}kA^{2} = \tfrac{1}{2}(50)(0.04) = 1.0\;\text{J}$.

At the center, all of that is kinetic: $\tfrac{1}{2}m v_{\max}^{2} = \tfrac{1}{2}(2)(1) = 1.0\;\text{J}$. ✓

Step 6 — A surprise: $\omega$ doesn't depend on amplitude.

If you doubled the amplitude (released from $x = 0.4\;\text{m}$ instead), the period would be the same $T \approx 1.257\;\text{s}$. The mass moves twice as fast at the center but has twice as far to go — those exactly cancel.

This isochronism is what made pendulum clocks possible: a clock's accuracy doesn't depend on whether you wind it tightly or loosely.

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Answer:

• $\omega = \sqrt{k/m} = 5\;\text{rad/s}$
• $T = 2\pi/\omega \approx 1.257\;\text{s}$
• $f = 1/T \approx 0.796\;\text{Hz}$
• $v_{\max} = A\omega = 1.000\;\text{m/s}$
• $a_{\max} = A\omega^{2} = 5.000\;\text{m/s}^{2}$

The mass oscillates back and forth, completing one full cycle every 1.257 seconds. Position, velocity, and acceleration are all sinusoidal but shifted in phase from each other.

Try It

• Adjust the mass, spring constant, and amplitude sliders.
• Watch the spring physically oscillate and the three sinusoidal traces (position cyan, velocity pink, acceleration yellow) drawn beside it.
• Notice that changing the amplitude does not change the period — that's the SHM signature.
• Doubling k with same m shrinks the period by $\sqrt{2}$.
• Doubling m with same k stretches the period by $\sqrt{2}$.