Questionnaire 1.2.5 — Oscillations

Four questions on the simple harmonic oscillator: computing the characteristic frequency (and not confusing Hz with rad/s), what "ideal" means for the amplitude, how the frequency scales with mass, and whether gravity has any say in a spring's frequency.

Key Ideas

The natural angular frequency of a mass on an ideal spring is

\omega_0 = \sqrt{\frac{k}{m}}

in radians per second. The frequency in hertz (cycles per second) is $f = \omega_0/2\pi$ , because one full cycle is $2\pi$ radians of phase — quoting $\omega_0$ when asked for hertz is the classic off-by- $2\pi$ error. The period is $T = 1/f$ .

An ideal oscillator has a restoring force exactly proportional to displacement ( $F = -kx$ ) and no friction of any kind. Its energy $\tfrac{1}{2}kA^2$ has nowhere to go, so the amplitude $A$ of an ideal oscillator stays constant forever — it does not decay at all, at any rate.

A real, linearly damped oscillator loses amplitude through the exponential envelope $A(t) = A_0\,e^{-\gamma t/2}$ : equal time intervals remove equal fractions of the amplitude, never equal amounts, and the envelope approaches zero without ever hitting it. "Decays at a constant rate" is wrong for a damped oscillator and doubly wrong for an ideal one.

Frequency scaling: $f \propto \sqrt{k/m}$ , so the mass enters under a square root — quadrupling $m$ halves $f$ ; quadrupling $k$ doubles it. Linear intuition ("4× the mass → ¼ the frequency") overshoots.

Gravity does not set a spring's frequency. A uniform gravitational field adds a constant force $mg$ , which only shifts the equilibrium point by $x_0 = mg/k$ ; the oscillation about that new equilibrium runs at the same $\omega_0 = \sqrt{k/m}$ . Contrast the pendulum, whose restoring force is gravity: $\omega = \sqrt{g/L}$ does change on the Moon.

Exercises

E1 (medium). Consider a mass hanging from an ideal spring. (An ideal spring will have restoring force proportional to how much it is stretched or compressed, and will have no friction. In such an ideal mass and spring problem, we will also presume there are no other sources of friction, such as air resistance on the mass.) Assume the mass is equal to $1\ \text{kg}$ and the spring constant is $10\ \text{N/m}$ . What is the characteristic frequency of this simple harmonic oscillator? That is, at approximately what frequency (in Hz or cycles per second) will the mass oscillate on the spring?

a) $1$ b) $0.503$ c) $3.16$ d) $10$ e) $19.9$ f) none of the above is close to the correct answer

Solution

Answer: b) $0.503$ .

Why, step by step:

The angular frequency of a mass-spring oscillator is $\omega_0 = \sqrt{k/m} = \sqrt{10/1} = \sqrt{10} \approx 3.16\ \text{rad/s}$ .
The question asks for hertz — cycles per second, not radians per second. One cycle is $2\pi$ radians, so $f = \dfrac{\omega_0}{2\pi} = \dfrac{3.16}{6.28} \approx 0.503\ \text{Hz}$ .
Sanity check: $f \approx 0.5\ \text{Hz}$ means a period of about two seconds — a lazy, very believable bounce for a 1 kg mass on a soft 10 N/m spring.

Why the tempting options fail:

c) $3.16$ is $\omega_0$ itself — correct number, wrong unit: it is the answer in rad/s, not Hz. This is the trap the question is built around.
e) $19.9 = 2\pi\sqrt{10}$ — multiplying by $2\pi$ instead of dividing.
a) $1 \approx \sqrt{10}/\pi$ — dividing by $\pi$ instead of $2\pi$ (half-remembered conversion).
d) $10$ is just $k/m$ with the square root forgotten.
f) tempts anyone who computed 3.16 rad/s, then (rightly) sensed the units were off but didn't know the fix.

See it: every wrong option is a wrong turn on this one-road derivation.

flowchart TD
    given["k = 10 N/m, m = 1 kg"] -->|"ω₀ = √(k/m)"| om["ω₀ = √10 ≈ 3.16 rad/s"]
    given -.->|"forget the √"| d10["10 ✗ — that's just k/m"]
    om -->|"÷ 2π rad per cycle"| ans["f ≈ 0.503 Hz ✓"]
    om -.->|"report as-is"| d316["3.16 ✗ — rad/s, not Hz"]
    om -.->|"× 2π instead"| d199["19.9 ✗"]
    om -.->|"÷ π instead"| d1["≈ 1 ✗"]

E2 (easy). True or false: consider the ideal mass-spring system from the first problem. The oscillator is given a finite initial kick to get it going. Given enough time, the oscillation amplitude will decrease at a constant rate until it reaches zero.

Solution

Answer: False.

Why, step by step:

"Ideal" was defined to mean no friction anywhere — no spring friction, no air resistance. Friction is the only way this oscillator can lose energy.
With nowhere for the energy $\tfrac{1}{2}kA^2$ to go, the amplitude $A$ stays exactly constant: the ideal oscillator rings forever. It does not decay at any rate, let alone a constant one.
The statement is even wrong about how real oscillators die: linear damping gives an exponential envelope $A_0 e^{-\gamma t/2}$ — equal times remove equal fractions, the per-second loss keeps shrinking, and the amplitude approaches zero without ever reaching it. "Constant rate until zero" describes neither the ideal nor the damped case.

Why the tempting reading fails:

Everyday springs and swings always die out, so "given enough time it stops" feels inevitable — but that intuition smuggles friction back into a problem that explicitly removed it.

See it: with the damping slider at $\gamma = 0$ the amplitude envelope is a flat line — give it all the time you want. Nudge $\gamma$ up and the envelope that appears is an exponential curve, not a straight ramp to zero.

E3 (medium). Consider the mass-spring system from the first problem. If I make the mass four times larger, does the frequency:

a) stay the same b) decrease to $1/4$ of its original value c) decrease to $1/2$ of its original value d) increase to $2$ times its original value e) increase to $4$ times its original value f) none of the above

Solution

Answer: c) decrease to $1/2$ of its original value.

Why, step by step:

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ , so the mass sits under a square root in the denominator.
Replace $m$ with $4m$ : $f' = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{4m}} = \dfrac{1}{\sqrt{4}}\cdot\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} = \dfrac{f}{2}$ .
Physically: more inertia, same restoring force per metre → the mass responds more sluggishly, so the frequency drops — but only by the square root of the mass factor. With the first problem's numbers, $0.503\ \text{Hz} \to 0.251\ \text{Hz}$ .

Why the tempting options fail:

b) $1/4$ is linear thinking — "four times the mass, a quarter the frequency" — which skips the square root that turns the factor 4 into a factor 2.
d), e) flip the direction: heavier must mean slower, not faster; $m$ is in the denominator.
a) confuses this with the (true, but different) fact that frequency doesn't depend on amplitude — it certainly depends on mass.

See it: drag the mass slider from 1 kg to 4 kg and watch the frequency readout go $0.503 \to 0.251$ Hz — half, not a quarter, because $f \propto 1/\sqrt{m}$ .

E4 (medium). Consider the mass-spring system from the first problem. Does it matter if the mass and spring are on the earth or on the moon? That is, does the frequency of oscillation change? (You can presume in both cases that the displacements of the mass are sufficiently small in height from the surface of the earth or the moon that the gravitational force on the mass does not depend on its position.)

a) yes b) no c) cannot be determined without more information

Solution

Answer: b) no.

Why, step by step:

Write Newton's law for the hanging mass with displacement $x$ measured from wherever the spring is unstretched: $m\ddot{x} = -kx + mg$ .
A constant extra force can't change the frequency — it only moves the equilibrium. Setting $x = x_0 + u$ with $x_0 = mg/k$ (the sag) gives $m\ddot{u} = -ku$ : exactly the gravity-free equation.
So on the Moon the spring sags about six times less, but the oscillation about that new equilibrium still runs at $\omega_0 = \sqrt{k/m}$ — $g$ appears nowhere in it. The frequency is set by the slope of the restoring force ( $k$ ), and uniform gravity adds a constant, not a slope.
The "small displacement" hint in the problem is what licenses treating $mg$ as constant — which is precisely the condition under which it drops out.

Why the tempting options fail:

a) borrows pendulum intuition: a pendulum's restoring force is gravity, so its $\omega = \sqrt{g/L}$ genuinely changes on the Moon. The spring's restoring force is the spring, not gravity.
c) hedges as if the answer hinged on unknown details ( $g_{\text{moon}}$ , the sag, the orientation) — but the derivation above eliminates $g$ entirely, no extra information needed.

See it: the spring and the pendulum start out ticking together on Earth. Drag $g$ down to the Moon's $1.62$ and watch which one falls out of step — and which only sags less.