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Physics · Ch 13 — Oscillations

Force Law for Simple Harmonic Motion

13.6

Force Law for Simple Harmonic Motion

The Connection Between Force and Displacement in SHM

Simple harmonic motion is not just a pattern of motion — it is defined by a specific relationship between the force acting on a particle and its displacement from equilibrium. The textbook section 13.6 establishes this force law, which is the physical cause behind the sinusoidal motion we have already studied.

When a particle executes SHM, its acceleration at any instant is given by a(t)=−ω2x(t)a(t) = -\omega^2 x(t), where x(t)x(t) is the displacement from the mean position. From Newton's second law, F=maF = ma, the force acting on the particle must be:

F=ma=m(−ω2x)=−mω2xF = m a = m(-\omega^2 x) = -m\omega^2 x

Since mm and ω\omega are constants for a given system, we can write k=mω2k = m\omega^2, where kk is a positive constant. This gives the force law for SHM:

F=−kxF = -k x

The negative sign is crucial: it tells us that the force always points towards the equilibrium position. When the particle is displaced to the right (x>0x > 0), the force is to the left (F<0F < 0), pulling it back. When displaced to the left (x<0x < 0), the force is to the right (F>0F > 0). This is why the force is called a restoring force — it always acts to restore the particle to its equilibrium position.

Important

The force in SHM is directly proportional to the displacement from equilibrium and opposite in direction to it. This is the defining physical condition for simple harmonic motion.

The Spring-Block System: A Concrete Example

The simplest physical system that obeys this force law is a block attached to a spring. For an ideal spring that obeys Hooke's law, the restoring force when the spring is stretched or compressed by a distance xx from its natural length is:

F=−ksxF = -k_s x

where ksk_s is the spring constant (or force constant) of the spring. Comparing this with F=−kxF = -kx, we see that for a spring-block system, k=ksk = k_s. The angular frequency of oscillation is then:

ω=km=ksm\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{k_s}{m}}

and the time period is:

T=2πω=2πmksT = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k_s}}

Watch out

Do not confuse the kk in F=−kxF = -kx with the spring constant ksk_s. In the general force law, k=mω2k = m\omega^2 is a constant that depends on both the system's mass and its stiffness. Only for a spring does kk equal the spring constant directly.

Properties of the Force Law

The textbook lists three important properties that follow from F=−kxF = -kx. Each one is derived directly from this equation.

›Proof

Property (I): The force is always directed towards the equilibrium position.

Let the equilibrium position be x=0x = 0. For any displacement xx:

  • If x>0x > 0 (particle to the right), then F=−kx<0F = -k x < 0, so force points left (towards x=0x=0).
  • If x<0x < 0 (particle to the left), then F=−kx>0F = -k x > 0, so force points right (towards x=0x=0).
  • If x=0x = 0, then F=0F = 0.

In every case, the force vector points from the particle's position back to the origin. This is the restoring nature of the force.

›Proof

Property (II): The force is proportional to the displacement.

From F=−kxF = -kx, the magnitude of the force is ∣F∣=k∣x∣|F| = k|x|. Doubling the displacement doubles the force magnitude; halving the displacement halves it. This linear relationship is what makes the motion "simple" — the differential equation it produces has sinusoidal solutions. If the force were proportional to x2x^2 or x\sqrt{x}, the motion would not be simple harmonic.

›Proof

Property (III): The constant kk determines the "stiffness" of the system.

For a given displacement xx, a larger kk means a larger restoring force. This makes the system "stiffer" — it resists displacement more strongly. Since ω=k/m\omega = \sqrt{k/m}, a larger kk gives a higher angular frequency and shorter time period. A smaller kk gives a weaker restoring force, lower frequency, and longer period. The constant kk is called the force constant or spring constant of the system.

The Differential Equation of SHM

Combining the force law with Newton's second law gives the equation of motion. Starting from F=maF = ma and F=−kxF = -kx:

md2xdt2=−kxm \frac{d^2 x}{dt^2} = -k x

Rearranging:

d2xdt2+kmx=0\frac{d^2 x}{dt^2} + \frac{k}{m} x = 0

Substituting ω2=k/m\omega^2 = k/m gives the standard differential equation for SHM:

d2xdt2+ω2x=0\frac{d^2 x}{dt^2} + \omega^2 x = 0 …