The Spring-Mass System: Why a Weight on a Spring Oscillates
Imagine hanging a weight from a spring and giving it a gentle tug downward. It bounces back up, overshoots, comes down again, and keeps going. That rhythmic up-and-down motion is simple harmonic motion, and the time it takes to complete one full bounce — down, up, and back to the start — is called the periodT.
The key question: what determines how fast or slow this bouncing happens? Intuition says two things matter: how heavy the weight is, and how stiff the spring is.
The Intuition
A heavier mass is harder to accelerate. It lumbers along, so the oscillation is slower — the period gets longer. A stiffer spring (one with a larger spring constant k) pulls back harder for the same stretch. That stronger restoring force whips the mass back faster, so the period gets shorter.
So the period T should increase with mass m and decrease with stiffness k. The exact relationship turns out to be:
T=2πkm
The 2π factor comes from the geometry of circular motion (which underlies all simple harmonic motion), and the square root tells us that doubling the mass only multiplies the period by 2≈1.4, not by 2.
Where Does This Formula Come From?
For a mass on an ideal spring, the restoring force is Hooke's law: F=−kx, where x is the displacement from equilibrium. Newton's second law F=ma gives:
mdt2d2x=−kx
This is a differential equation whose solution is a sine or cosine wave. The angular frequency ω (how fast the oscillation goes in radians per second) comes out as:
ω=mk
Since period T is the time for one complete cycle, and one cycle corresponds to 2π radians, we have T=2π/ω. Substituting ω gives the formula above.
Tip
If you ever forget which variable goes in the numerator, remember: mass in the numerator makes the period longer (heavier = slower), and stiffness in the denominator makes the period shorter (stiffer = faster). The square root just moderates the effect.
What the Formula Tells You
Mass and period: Double the mass → period increases by 2 (about 1.4 times). Quadruple the mass → period doubles.
Stiffness and period: Double the spring constant → period decreases by 2 (about 0.7 times). Quadruple the stiffness → period halves.
Independence from amplitude: The period does not depend on how far you pull the mass initially. A small bounce and a big bounce take exactly the same time. This is the hallmark of simple harmonic motion for an ideal spring. …
A spring pendulum (mass m on a spring of constant k) oscillates with T=2πm/k; since this expression contains no g at all, the period is independent of the acceleration due to …
A mass-spring system's period, T=2πm/k, has no g in it at all, so it is the same everywhere regardless of local gravity.
A spring pendulum is a system consisting of a point mass m attached to one end of a massless, elastic spring of force constant k, free to oscillate (e.g. hanging vertically or on a frictionless horizontal surface).
When the mass hangs in equilibrium, the spring stretches by an amount x0 such that the spring force balances gravity: kx0=mg.
If the mass is displaced by a further small distance x from this equilibrium and released, the net restoring force is F=−k(x0+x)+mg=−kx0−kx+mg=−kx (since kx0=mg cancels the weight). So the net restoring force about the equilibrium position is simply F=−kx, independent of g.
By Newton's second law, mdt2d2x=−kx, i.e. dt2d2x=−mkx=−ω2x, which is the SHM equation with angular frequency ω=k/m.
Q.What will be the change in time period of a spring pendulum when taken to the moon?
›Reveal solutionSolution
A spring (mass-spring system) pendulum's restoring force comes from the spring constant, not from gravity, so its period is independent of the local value of g — unlike a simple pendulum.
For a mass m attached to a spring of force constant k, oscillating horizontally or vertically, the restoring force is F=−kx, and Newton's second law gives:
T=2πkm
This expression contains only the mass m and the spring constant k; it does not contain the acceleration due to gravity g at all (gravity only shifts the equilibrium position for a vertical spring, it does not change the restoring-force constant).
Q.A mass-spring system is oscillating in a car. If the car moves on a horizontal road with accelerated speed, then its frequency
(A) will decrease
(B) will increase
(C) will remain same
(D) will be zero
›Reveal solutionSolution
A mass-spring system's oscillation frequency is unaffected by the accelerating frame it's placed in.
The frequency of a spring-mass oscillator is f=2π1mk, which depends only on the spring constant k and the oscillating mass m. If the car accelerates, a pseudo-force acts on the system in its (non-inertial) frame, but this only shifts the equilibrium (mean) posit …
Q.Two bodies A and B whose masses are in the ratio 1 : 2 are suspended from two separate massless springs of force constants kA and kB respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio 1 : 2, the ratio of the amplitude A to that of B is ____.
(a) sqrt(2 kB / kA)
(b) sqrt(kB / 2 kA)
(c) sqrt(8 kB / kA)
(d) sqrt(kB / 8 kA)
›Reveal solutionSolution
Using v_max = A√(k/m) for each body and the given mass ratio (1:2) and velocity ratio (1:2), the amplitude ratio works out to sqrt(kB/8kA).
For a mass m attached to a spring of constant k, undergoing SHM with amplitude A, the maximum velocity is:
v_max = A ω = A √(k/m)
Let mass of A be m, so mass of B = 2m (given mA : mB = 1 : 2).
Given v_maxA : v_maxB = 1 : 2, i.e. v_maxA / v_maxB = 1/2.
The spring constant k, defined by F = -kx, has SI unit N/m (equivalently kg/s²).
Hooke's law for a spring states that the restoring force is F = -kx, where x is the displacement from the natural length and k is the spring (force) constant. Rearranging, k = -F/x, so its unit is [Force]/[Length] = N/m. This same constant …