Unit 7 · Topic 7.3

7.3 Representing and Analyzing SHM

Newton's second law for SHM gives the equation $\dfrac{d^2x}{dt^2} = -\omega^2 x$ , and its solution is $x(t) = A\cos(\omega t + \phi)$ . Differentiating gives the velocity and acceleration, with maximum speed Aω and maximum acceleration Aω². Driving a system at its natural frequency causes resonance.

Key terms

amplitude
phase constant
second-order differential equation
natural frequency
resonance

What changed in the 2024 update

Before the Fall 2024 course update, oscillations were Unit 6. They're now Unit 7, so older videos and practice may call this unit "Unit 6."

The equation of SHM

For a mass on a spring, Newton's second law gives $m\dfrac{d^2x}{dt^2} = -kx$ . Divide by m: $\frac{d^2x}{dt^2} = -\omega^2 x \qquad \text{with } \omega = \sqrt{\frac{k}{m}}$ This is a second-order differential equation: it relates a function to its second derivative.

Its solution is $x(t) = A\cos(\omega t + \phi)$ where A is the amplitude and φ is the phase constant, which sets where in the cycle the object starts. You can check it: differentiating cos twice gives back −ω² times the same function.

You need to know this solution and recognize the SHM equation, but you won't have to solve differential equations in general. Any system whose equation fits this form is SHM, with ω equal to the square root of the constant.

Velocity and acceleration

Differentiate the position: $v(t) = -A\omega\sin(\omega t + \phi) \qquad a(t) = -A\omega^2\cos(\omega t + \phi) = -\omega^2 x$

So the maximum speed is $v_{\max} = A\omega$ and the maximum acceleration is $a_{\max} = A\omega^2$ . Both grow with the amplitude, even though the period doesn't.

The phase constant comes from the starting conditions. Released from rest at x = +A: φ = 0, so x = A cos ωt. Starting at equilibrium moving in the + direction: x = A sin ωt, which is the same as φ = −π/2.

Reading SHM graphs

The x, v and a graphs are all sinusoids with the same period, shifted from each other by a quarter cycle. The acceleration graph is the position graph flipped upside down, because a = −ω²x.

Velocity is the slope of the position graph: zero at the peaks and troughs, biggest where the position graph crosses zero.

Where	Position	Speed	Acceleration and net force
turning points (x = ±A)	maximum size	zero	maximum size, toward equilibrium
equilibrium (x = 0)	zero	maximum	zero

Resonance

Every oscillating system has a natural frequency: the frequency it oscillates at when you displace it and let go. For a spring–mass system, that's (1/2π)√(k/m).

If an outside force pushes back and forth on the system at its natural frequency, each push adds energy at the right moment, and the amplitude grows. This is resonance. Pushing a child on a swing in time with the swing is the everyday example. Pushing at other frequencies gives much smaller amplitudes.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Reading an SHM equation
An object's position is $x(t) = (0.050\text{ m})\cos(8.0t + \pi/3)$ , with t in seconds. Find the amplitude, angular frequency, period, maximum speed and maximum acceleration. Then find the position and velocity at t = 0.
Show the solutionHide the solution
1. Step 1: Read off A = 0.050 m and ω = 8.0 rad/s. Then $T = \dfrac{2\pi}{8.0} \approx 0.79$ s.
2. Step 2: $v_{\max} = A\omega = (0.050)(8.0) = 0.40$ m/s and $a_{\max} = A\omega^2 = (0.050)(64) = 3.2$ m/s².
3. Step 3: At t = 0: $x = 0.050\cos(\pi/3) = 0.025$ m.
4. Step 4: $v = -A\omega\sin(\omega t + \phi)$ , so $v(0) = -(0.40)\sin(\pi/3) \approx -0.35$ m/s. The object is halfway out on the + side, moving back toward equilibrium.
Answer: A = 0.050 m, ω = 8.0 rad/s, T ≈ 0.79 s, v_max = 0.40 m/s, a_max = 3.2 m/s²; x(0) = 0.025 m, v(0) ≈ −0.35 m/s
Example 2Calculator allowed
Writing x(t) from starting conditions
A 0.25 kg block on a spring (k = 100 N/m) passes through equilibrium at t = 0, moving in the +x direction at 0.60 m/s. Write x(t).
Show the solutionHide the solution
1. Step 1: $\omega = \sqrt{\dfrac{100}{0.25}} = 20$ rad/s.
2. Step 2: Equilibrium is where the speed is greatest, so $v_{\max} = A\omega = 0.60$ m/s and $A = \dfrac{0.60}{20} = 0.030$ m.
3. Step 3: It starts at x = 0 moving in the + direction, so the sine form fits: x = A sin ωt. Check: x(0) = 0, and v(0) = Aω cos 0 = +0.60 m/s.
4. Step 4: So $x(t) = (0.030\text{ m})\sin(20t)$ , which is the same as $0.030\cos(20t - \pi/2)$ .
Answer: x(t) = 0.030 sin(20t), in meters
Example 3
Speed and acceleration at the same moment (classic trap)
A student says: "The block on the spring is fastest where its acceleration is largest, because acceleration makes things go faster." Is the student right?
Show the solutionHide the solution
1. Step 1: No. The acceleration is largest at the turning points, x = ±A, where a = −ω²x has its largest size. The speed there is zero.
2. Step 2: At equilibrium, x = 0, so the acceleration is zero, and that's where the speed is greatest.
3. Step 3: Large acceleration means the velocity is changing quickly, not that it's large. At the turning points the velocity is reversing as fast as it ever does.
Answer: No: the acceleration is largest where the speed is zero, and zero where the speed is largest

Common mistakes

Using degrees in cos(ωt + φ). ωt is in radians, so set your calculator to radian mode.
Thinking v_max or a_max is independent of amplitude because the period is. Both are proportional to A.
Writing the velocity as +Aω sin(ωt + φ). The derivative of cos brings a minus sign.
Thinking resonance means any large push. It means pushing at the system's natural frequency.

On the exam

Translation Between Representations questions often give one of the x, v or a graphs and ask you to sketch the others on the same time axis. Line up the zeros and peaks carefully.
To derive ω for a new system, write Newton's second law, rearrange to d²x/dt² = −(constant)x, and identify ω² as the constant.

Connected topics

Videos

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