Before any trap makes sense, we need three points pinned down on one axis and a sign convention. Look at the figure: a hanging spring is drawn from a hatched ceiling; the red block is the mass m, sitting at rest. Two horizontal dashed lines mark the reference heights — the upper dashed line is where the spring's free end would sit with no mass (natural length), and the lower dashed line is where the mass actually rests (equilibrium). The black double-headed arrow on the left measures the gap between them, labelled y0. The downward black arrow on the right marks the positive x direction and shows that x=0 is placed at equilibrium, not at the natural length.
Recall Where
T=2πm/k actually comes from (full derivation, so the page stands alone)
Start from the restoring force measured from equilibrium, F=−kx, and Newton's law F=mx¨:
mx¨=−kx⇒x¨=−mkx.
This says: acceleration is a fixed negative multiple of position. Call that multiple ω2, i.e. define ω2=k/m, so x¨=−ω2x.
Now guess a wobble x(t)=Acos(ωt+ϕ) (a cosine is the only shape whose second derivative is a negative copy of itself). Differentiate twice:
x˙=−Aωsin(ωt+ϕ),x¨=−Aω2cos(ωt+ϕ)=−ω2x.✓
It fits for any amplitude A and phase ϕ, confirming ω=k/m.
A cosine repeats when its angle advances by 2π: ωT=2π, so
T=ω2π=2πkm.
That is the formula every trap on this page probes.
Recall Why the gravity term cancels (re-derived here too)
Measure raw position y downward from the natural length (upper dashed line). Two forces act: gravity +mg (down) and spring −ky (up, proportional to stretch). Newton:
my¨=mg−ky.
Now switch to the from-equilibrium coordinate x=y−y0 with y0=mg/k. Since y0 is a constant, y¨=x¨, and:
mx¨=mg−k(x+y0)==0(mg−ky0)−kx=−kx.
The constant mg was fully absorbed by the shift — nothing g-dependent survives in the dynamics. The figure above shows exactly this shift: gravity relocated the origin from the upper dashed line down to the lower one, and about that new origin the physics is identical to the horizontal case.
False — T∝m, so Tnew/Told=2m/m=2≈1.41, not 2. The square root is the whole trap.
Doubling the stiffness k halves the period.
False — T∝1/k, so Tnew/Told=k/2k=1/2≈0.71. To halveT you'd need to quadruple k.
A vertical spring-mass system oscillates more slowly than the same spring horizontally.
False — both give T=2πm/k. Gravity is constant, so it only relocates the rest point; the restoring term about the new equilibrium is still −kx (see the re-derivation box above).
Increasing the amplitude A increases the period of an ideal spring-mass system.
False — from the derivation box, the solution x(t)=Acos(ωt+ϕ) has ω=k/m with no A inside, so T=2πm/k is amplitude-independent ("isochronous"). Physically, scaling x→cx scales x¨=−(k/m)x by the same factor c, so the whole wobble is just taller with identical timing.
On the Moon, a horizontal spring-mass oscillator has the same period as on Earth.
True — the horizontal period T=2πm/k contains no g at all, so gravity strength is irrelevant.
On the Moon, a vertical spring stretches less at rest but keeps the same period as on Earth.
True — static extension y0=mg/k shrinks with weaker g, but the dynamics still reduce to mx¨=−kx, so T is unchanged.
At the equilibrium point the net force is zero, so the mass momentarily stops there.
False — net force is zero at equilibrium, but that's exactly where speed is maximum. Zero force means zero acceleration, not zero velocity.
At the extreme positions (turning points, x=±A) the mass has zero speed but non-zero acceleration.
True — speed vanishes as it reverses, while displacement is largest so ∣F∣=k∣x∣ and acceleration are at their peak.
The formula T=2πy0/g shows the vertical period depends on gravity.
False — g appears but only because y0=mg/k hides the m/k; substituting y0=mg/k gives 2π(mg/k)/g=2πm/k and the g cancels. It's a measurement shortcut, not a g-dependence.
Two identical springs in parallel give a longer period than one spring alone.
False — parallel doubles the effective stiffness (keff=2k; see the note below), and stiffer means faster, so the period shortens by 2: T=2πm/2k.
The three "wrong equations" below all live on the same axis as the figure at the top — the fix is always about which reference point x is measured from.
"For the hanging spring I measure x from the natural (unstretched) length, then mx¨=−kx."
Wrong reference point — from the natural length the equation reads mx¨=mg−kx, which still carries a leftover mg. You must measure from equilibrium (the point a distance y0=mg/k below the natural length) for the constant to cancel.
"Gravity adds a downward force, so the vertical equation is mx¨=−kx+mg and gravity clearly affects the motion."
The +mg is real only while x is measured from the natural length; shifting the origin down to equilibrium absorbs it into a constant that cancels (ky0=mg), leaving the oscillation term purely −kx.
"Heavier mass moves more, so a heavier block on the same spring oscillates faster."
Backwards — T∝m, so more mass means more inertia resisting the spring and a slower (longer-period) oscillation.
"Since F=−kx, the acceleration is constant like a falling body, so we can use v2=u2+2as (with u the initial speed, v the final speed, s the distance)."
No — those constant-acceleration formulas assume a is fixed, but here a=−kx/mchanges with position because it depends on x. Constant-acceleration kinematics never apply to SHM; use energy conservation or x(t)=Acos(ωt+ϕ).
"The minus sign in F=−kx just means the spring pushes; I can drop it and still get T=2πm/k."
The minus sign is the whole physics — it makes the force restoring (points back to equilibrium). Drop it and mx¨=+kx gives runaway exponential growth (x∼ek/mt), not oscillation.
"Springs in series add up like keff=k1+k2, so series is stiffer."
That's the parallel rule. In series the stretches add for the same force, giving keff1=k11+k21, which is softer than either spring, so series oscillates slower.
Why does the period not depend on amplitude, unlike your gut feeling?
Write the equation of motion x¨=−(k/m)x: the acceleration is a fixed multiple−k/m of the position, so scaling the whole motion x→cx scales x¨→cx¨ identically — the same sinusoid, just taller. Formally x(t)=Acos(ωt+ϕ) has ω=k/m with no A inside, so the period is fixed for any A.
Why must the restoring force be proportional to displacement (not just opposing it) for true SHM?
Only a strictly linear −kx produces the equation x¨=−ω2x whose solution is a pure sinusoid with a single, amplitude-independent frequency. Any other restoring law gives anharmonic motion whose period drifts with amplitude.
Why can a stiffer spring make the oscillation faster even though it's "harder to move"?
Stiffness k sits under the square root as T∝1/k: a stiffer spring yanks the mass back with more force per unit displacement, so it accelerates home quicker, shortening the period.
Why does gravity move the equilibrium position but not the value of the period?
Gravity is a constant force, so it only sets a new balance point where ky0=mg; oscillation is driven by the change in spring force as you move from that point, which is still −kx, giving the same T=2πm/k.
Why does energy conservation give the same ω as Newton's law?
Differentiating E=21mv2+21kx2 and setting dE/dt=0 factors into v(mx¨+kx)=0; since v=0 generally, you recover mx¨+kx=0 — the identical SHM equation, hence the same ω=k/m.
Why is the speed maximum exactly at the centre and zero at the ends?
All the energy is a trade between spring PE (21kx2) and KE (21mv2). At the centre x=0 so PE is zero and KE holds the whole budget (max speed); at the extremes x=±A so PE is maximal and KE, and speed, drop to zero.
What happens to the period if the mass is (idealised as) zero?
T=2πm/k→0 — a massless block would oscillate infinitely fast. This is unphysical and flags that a real spring's own mass would then dominate.
What happens as k→0 (an infinitely floppy spring)?
T=2πm/k→∞ — with vanishing restoring force there's almost nothing pulling the mass back, so the "oscillation" takes forever. In the limit it stops being an oscillator.
If a horizontal spring has friction on the surface, is the motion still SHM with this period?
No — friction adds a velocity-dependent force, turning it into a damped oscillator; the amplitude decays and the period lengthens slightly from the ideal 2πm/k.
Is there any oscillation if the mass starts exactly at equilibrium with zero velocity?
No — net force and velocity are both zero, so it simply stays put (amplitude A=0). Oscillation needs either an initial displacement or an initial push.
Compare the spring's period to a pendulum's at a degenerate limit: which one is affected by moving to another planet?
Only the pendulum (T=2πL/g) changes with g; the spring's T=2πm/k carries no g, so it ticks identically anywhere.
If a vertical spring is stretched so far it goes slack on the way up (spring can only pull), is the motion still pure SHM?
No — once the spring goes slack the force is no longer −kx (it's just gravity/free flight for part of the cycle), so the linear restoring law breaks and the motion is no longer sinusoidal.
Recall One-line self-test (formal version)
Fill in with physical quantities: "gravity only shifts the ___, never the ___." ::: the equilibrium position (y0=mg/k) ; the periodT=2πm/k.
