Force law for SHM

What is the force law for SHM?

Like acceleration, force is always directed towards the mean position—hence it is sometimes called the restoring force in SHM.

NCERT Physics Part II (Class 11), Section 13.6, page 267

In plain words: pull the particle away from its resting place and something pulls it back, and the further you pull it, the harder it pulls back, in exact proportion. With constant mass and that net force, a displaced or moving particle executes SHM, and the stiffer the pull (larger k) or the lighter the particle (smaller m), the faster it oscillates.

From displacement to force

The previous section of the chapter showed that a particle in SHM has acceleration a(t) = −ω^²x(t) (NCERT Eq. 13.11). Newton’s second law then gives the force on it:

So SHM can be defined in two equivalent ways (page 267):

• by its displacement, x(t) = A cos(ωt + φ) (NCERT Eq. 13.4); or
• by its force law, F = −kx (NCERT Eq. 13.13).

Differentiating the displacement twice gives the acceleration, and multiplying by m gives the force law. Going the other way, solving the differential equation m d^²x/dt^² = −kx gives the displacement equation, with A and φ set by how the motion starts.

A quick test for SHM. If a problem gives you an acceleration of the form a = −(a positive constant) × x, the motion is SHM and that constant is ω^² (Example 3). A constant term only moves the mean position: a = −ω^²(x − x₀) is SHM about x = x₀ with the same ω (the vertical spring below is an example).

Restoring force and force constant

The force always points back towards the mean position, so it is called the restoring force. The force constant k has the unit N m^⁻¹: it is the restoring force per metre of displacement.

Because F is proportional to the first power of x, a particle moving under it is called a linear harmonic oscillator. In real systems the force may also contain small extra terms proportional to x^², x^³ and so on; such systems are non-linear oscillators (page 267). When a nonzero positive linear restoring coefficient exists, higher powers become relatively small for sufficiently small displacements, which is why so many vibrations are very nearly simple harmonic.

Oscillations of a spring

A block of mass m on a smooth horizontal surface, fixed to a light (massless) spring of spring constant k, is the standard example. By Hooke’s law the spring pulls or pushes the block with force −kx when it is displaced by x, so the block moves in SHM with ω = √(k/m) and period

This follows from T = 2π/ω and NCERT Eq. 13.14b. Notice what is not in the formula: the amplitude. A spring–block oscillator takes the same time for a small swing as for a large one (within the range where Hooke’s law holds).

Two springs. NCERT Example 13.6 (pages 267–268) fixes a block between two identical springs, each of constant k. Displace the block by x: one spring stretches by x and the other compresses by x, and both forces point back towards the mean position. The net force is −2kx, so the motion is SHM with T = 2π√(m/2k) (Example 2).

Combinations of springs (beyond the NCERT text):

• in parallel (side by side, or on either side of the block as in NCERT Example 13.6), the magnitudes of their changes in length are the same and their restoring contributions add (opposite-side springs have opposite signed changes in length): k_eff = k₁ + k₂;
• in series (end to end), both carry the same force and their extensions add: 1/k_eff = 1/k₁ + 1/k₂, so k_eff = k₁k₂/(k₁ + k₂);
• cutting an ideal uniform spring, neglecting end effects and changes to its geometry: k is inversely proportional to its length, so each half of a spring of constant k has constant 2k.

A vertical spring (also beyond the NCERT text). Hanging a mass m on a spring stretches it by Δl = mg/k to a new equilibrium. Take x downward from that new equilibrium: the net force is mg − k(Δl + x) = −kx, because kΔl = mg. So the period is the same, T = 2π√(m/k) = 2π√(Δl/g) (Example 4).

Solved examples

Common mistakes to avoid

Quick revision summary

• Force law for SHM: F = −kx. The force is the restoring force, always towards the mean position.
• From a = −ω^²x and F = ma: k = mω^², ω = √(k/m).
• SHM can be defined either by x(t) = A cos(ωt + φ) or by F = −kx.
• A linear harmonic oscillator has F ∝ −x; extra x^², x^³ terms make it non-linear.
• Spring–block: T = 2π√(m/k), independent of amplitude; a block between two identical springs, each of constant k, has effective constant 2k.
• Springs in parallel: k = k₁ + k₂; in series: k = k₁k₂/(k₁ + k₂).

Questions students ask

Why does a spring’s period depend on the mass when a pendulum’s does not?

The period of a block on a spring, T = 2π√(m/k), depends on the mass because the force constant k is fixed by the spring alone: a heavier block gets the same restoring force for the same stretch but accelerates less. A pendulum’s restoring force, mg sin θ ≈ (mg/L)x for small swings (x = Lθ along the arc), is itself proportional to the mass, so its force constant mg/L grows with m and the mass cancels: T = 2π√(m/(mg/L)) = 2π√(L/g), as on the simple pendulum page.

What happens to the spring constant when a spring is cut into two unequal parts?

When a uniform spring is cut, the spring constant of each piece is inversely proportional to its length, so the shorter piece is the stiffer one: k₁l₁ = k₂l₂ = kl. A spring of constant 60 N m^⁻¹ cut in the ratio 1 : 2 gives pieces one-third and two-thirds as long, with constants 180 N m^⁻¹ and 90 N m^⁻¹; joined again in series they give back 1/k = 1/180 + 1/90 = 1/60, that is 60 N m^⁻¹. A block on the shorter piece oscillates with 1/√3 of its period on the whole spring.

What is the difference between Hooke’s law and the force law for SHM?

Hooke’s law is a property of a spring or elastic body, while the force law for SHM is a condition on the net force on any oscillating body, though both have the form F = −kx. Hooke’s law says that, for small deformations, the force a spring exerts is proportional to its extension or compression. The force law for SHM says that a body moves in SHM when the net force on it, whatever its source, is −kx: a pendulum bob’s tangential motion for small swings, a floating body bobbing gently (weight and buoyancy) and a block between two springs all qualify, and k need not be the constant of any one spring.

What is the period of two blocks joined by a spring on a smooth floor?

Two blocks of masses m₁ and m₂ joined by a light spring of constant k on a smooth floor oscillate with period T = 2π√(μ/k), where μ = m₁m₂/(m₁ + m₂) is the reduced mass (a step beyond the NCERT text). Relative to their uniformly moving centre of mass, both blocks execute SHM in opposite directions. In a laboratory frame with a moving centre of mass, the motion includes uniform drift and is not generally periodic. For 2 kg and 3 kg blocks on a 300 N m^⁻¹ spring, μ = 6/5 = 1.2 kg and T = 2π√(1.2/300) ≈ 0.40 s, shorter than for either block alone on the same spring fixed to a wall.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. What is a restoring force? Why is it called that?

   Show answerHide answer

   In SHM the force on the particle always points towards the mean position, F = −kx. It tends to restore the particle to its equilibrium position, hence the name.

2. Q2. A 2 kg block is attached to a spring of constant 450 N m^⁻¹. Find the angular frequency, period and frequency of its oscillations.

   Show answerHide answer

   ω = √(450/2) = 15 rad/s, T = 2π/15, about 0.419 s, and ν = 1/T, about 2.39 Hz.

3. Q3. The amplitude of a spring–block oscillator is doubled. What happens to its period?

   Show answerHide answer

   Nothing: T = 2π√(m/k) does not depend on the amplitude. The block simply moves faster over the longer path.