# 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:**
1. The angular frequency of a mass-spring oscillator is
$\omega_0 = \sqrt{k/m} = \sqrt{10/1} = \sqrt{10} \approx 3.16\ \text{rad/s}$.
2. The question asks for hertz — cycles per second, not radians per second. One cycle is
$2\pi$ radians, so
$f = \dfrac{\omega_0}{2\pi} = \dfrac{3.16}{6.28} \approx 0.503\ \text{Hz}$.
3. 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.
```mermaid
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 ✗"]
```
Solution
**Answer: False.**
**Why, step by step:**
1. "Ideal" was defined to mean _no friction anywhere_ — no spring friction, no air
resistance. Friction is the only way this oscillator can lose energy.
2. 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.
3. 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.
```qm-demo mass-spring-oscillation
```
Solution
**Answer: c) decrease to $1/2$ of its original value.**
**Why, step by step:**
1. $f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$, so the mass sits _under a square root_ in the
denominator.
2. Replace $m$ with $4m$:
$f' = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{4m}} = \dfrac{1}{\sqrt{4}}\cdot\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} = \dfrac{f}{2}$.
3. 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 $f \propto 1/\sqrt{m}$.
```qm-demo mass-spring-oscillation
```
Solution
**Answer: b) no.**
**Why, step by step:**
1. Write Newton's law for the hanging mass with displacement $x$ measured from wherever the
spring is unstretched: $m\ddot{x} = -kx + mg$.
2. 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.
3. 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.
4. 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.
```qm-demo hanging-spring-equilibrium
```