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Hang a mass on a spring, pull it down and let go. It bobs up and down, and gravity is pulling on it the whole time. Does gravity change the period? Many students think it must. It does not, and seeing why makes a whole family of JEE and NEET questions easy. This post builds the idea in three steps and works through amplitude, speed, force and the case where the spring goes slack.
What does the syllabus ask?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list oscillations of a spring, with restoring force and force constant, and energy in SHM. The NCERT Class 11 chapter on Oscillations derives F = −kx and ω = √(k/m). Its exercises include a body hanging from a spring balance that oscillates when pulled, which is exactly this set-up.
Where is the new equilibrium?
At rest, the spring force balances the weight: k x₀ = mg, so the static extension is
x₀ = mg/k.
This is the new centre of the motion. It is not the natural length of the spring.
Why is the period unchanged?
Measure displacement y downward from the new equilibrium. The spring is then stretched by x₀ + y, so the net downward force is
mg − k(x₀ + y) = mg − kx₀ − ky = −ky,
because mg = kx₀. Gravity and the extra spring force from x₀ cancel exactly, leaving the same restoring force as a horizontal spring. So the motion is SHM with
T = 2π√(m/k) = 2π√(x₀/g).
The second form is handy: if you know how far the spring stretches when the mass is hung, you know the period without knowing m or k.
Worked example 1: released from the natural length
A 0.50 kg mass is attached to a vertical spring with k = 50 N/m. It is held so the spring is unstretched, then released from rest. Take g = 10 m/s².
Static extension: x₀ = 0.50 × 10 / 50 = 0.10 m.
Angular frequency: ω = √(50/0.50) = 10 rad/s. Period T = 2π/10 ≈ 0.63 s. Check: 2π√(0.10/10) = 2π × 0.10, the same.
The mass starts at rest 0.10 m above equilibrium, so the amplitude is A = 0.10 m. The lowest point is 0.20 m below the natural length, which is 2mg/k.
Maximum speed, at equilibrium: Aω = 0.10 × 10 = 1.0 m/s.
Spring force at the lowest point: 50 × 0.20 = 10 N, twice the weight. At the top it is zero, and the acceleration there is ω²A = 100 × 0.10 = 10 m/s², which is just g, as it should be with only gravity acting.
Worked example 2: pulled down from equilibrium
The same mass hangs at rest. It is pulled down a further 0.05 m and released.
Amplitude A = 0.05 m; the period is still about 0.63 s.
The spring extension swings between 0.10 − 0.05 = 0.05 m and 0.10 + 0.05 = 0.15 m.
Spring force swings between 50 × 0.05 = 2.5 N and 50 × 0.15 = 7.5 N.
Maximum speed = 0.05 × 10 = 0.50 m/s.
Worked example 3: period from the stretch alone
A spring stretches by 2.5 cm when a mass is hung on it. Find the period, with g = 9.8 m/s².
T = 2π√(0.025/9.8) = 2π × 0.0505 ≈ 0.32 s.
No mass or spring constant needed.
When does the simple picture break?
The method needs the spring to stay stretched. In example 2, the extension never drops below 0.05 m, so it does. If the amplitude were larger than x₀ for a spring that cannot push, such as a rubber cord, the cord would go slack at the top and the motion would no longer be SHM throughout. For a coil spring that can be compressed, SHM continues past the natural length.
Energy works out neatly too. Measured from the new equilibrium, the spring's elastic energy and the gravitational energy together change exactly like ½ky², so the usual SHM energy results hold. Our guide to simple harmonic motion problems covers other SHM set-ups, and spring-block energy problems covers the drop-onto-a-spring version.
Which mistakes cost marks?
Taking the natural length as the mean position. The mean position is where the forces balance.
Thinking g changes the period. It changes x₀, and x₀/g = m/k is unchanged.
Confusing the static stretch with the lowest point. A mass lowered gently stops at mg/k; a mass dropped from the natural length goes to 2mg/k.
Double-counting gravity in energy. Use either spring energy plus gravitational energy from a fixed level, or ½ky² from the new equilibrium, not both.
Ignoring slack. Check that the amplitude is less than mg/k for a cord or string.
A practice routine
Derive T = 2π√(x₀/g) from a blank page.
Redo example 1 with m = 0.20 kg and k = 80 N/m.
Solve NCERT Exercise 13.8 on the spring balance, finding k from the scale first.
Solve one problem where a block sits on a mass on a vertical spring and find when it loses contact.
If you want to see the mean position shift as you add mass, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Next time you are near a spring, hang something on it, measure the stretch, and predict the period before you time it.
