Spring Simple Harmonic Motion Trig Problems

The general equation of simple harmonic motion for a spring is written as:

\( y = A\cos(bt) \)   or   \( y = A\sin(bt) \)

where:

• \(y\) is the displacement of the mass from its resting (equilibrium) position at time \(t\)
• \(A\) is the amplitude, the maximum distance the mass moves away from equilibrium
• \(b\) is the angular frequency, related to the period \(T\) (the time for one full bounce) by \( b = \dfrac{2\pi}{T} \)

These same building blocks, amplitude and period turning into a horizontal stretch or compression of the graph, are the ideas covered in transformations of trigonometric functions. A spring problem is simply that transformation idea applied to a physical motion instead of an abstract graph.

The only real decision in a spring problem is which function to start from, and that depends entirely on where the mass is at \(t = 0\):

• If the spring is pulled down (or pushed up) to its maximum stretch and released from rest, the motion starts at a maximum, so use \( y = A\cos(bt) \).
• If the spring is released while passing through its natural resting position, moving fastest at that instant, the motion starts at zero, so use \( y = A\sin(bt) \).

Sign matters too: if the mass starts below equilibrium and moves upward first, or starts above and moves downward first, you may need a negative amplitude, that is, \( y = -A\cos(bt) \) or \( y = -A\sin(bt) \), to match the actual direction of the first swing.

Spring positions at a glance

A mass is attached to a spring and pulled down 4 cm below its resting position, then released. It takes 6 seconds to complete one full bounce cycle. Write an equation for its displacement \(y\) (in cm) as a function of time \(t\) (in seconds), and find its position after 2 seconds.

Step 1: Identify the amplitude. The mass moves 4 cm from equilibrium, so \( A = 4 \).

Step 2: Identify the period and find \(b\). One full cycle takes \( T = 6 \) seconds, so \( b = \dfrac{2\pi}{T} = \dfrac{2\pi}{6} = \dfrac{\pi}{3} \).

Step 3: Choose sine or cosine. The mass starts at its maximum displacement (pulled down, released from rest), so the motion begins at an extreme, which matches a cosine curve. Since it starts below equilibrium, take that starting displacement as negative relative to the usual "up is positive" convention, or simply define down as positive from the start, whichever the problem intends. Here we treat downward displacement as positive, so:

\( y = 4\cos\left(\dfrac{\pi}{3}t\right) \)

Step 4: Evaluate at \(t = 2\).

\( y = 4\cos\left(\dfrac{\pi}{3}(2)\right) = 4\cos\left(\dfrac{2\pi}{3}\right) = 4(-0.5) = -2 \)

So after 2 seconds, the mass is at \(y = -2\) cm, meaning it has moved 2 cm past equilibrium in the opposite direction from where it started.

On the graph, the curve starts at its highest point \(y = 4\) when \(t = 0\), the classic starting shape of a cosine curve. It crosses \(y = 0\) a quarter of the way through the period, reaches its lowest point \(y = -4\) at the halfway mark, crosses zero again three-quarters of the way through, and returns to \(y = 4\) exactly one full period later. Every spring SHM graph follows this same rhythm; only the amplitude and period values change from problem to problem.

The exact same reasoning, amplitude, period, and choosing between sine and cosine based on the starting point, appears whenever something moves back and forth or up and down in a repeating pattern. You will see it again in tides and water depth trig problems, where water height rises and falls with a similar rhythm, and in Ferris wheel motion problems, which use a rotating version of the same idea. Once you can set up a spring's equation confidently, those related problems become a matter of relabeling the same four steps.

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