Mass-spring system

The ideal spring oscillator

A mass-spring system consists of a mass $m$ attached to a spring with spring constant $k$. In the ideal model, the spring is massless, obeys Hooke's law, and there is no friction or air resistance. The mass moves along one axis, and displacement $x$ is measured from equilibrium.

Hooke's law says the spring force is

$F_s=-kx$

The negative sign means the force points opposite the displacement. If the spring is stretched to positive $x$, the force pulls back toward equilibrium. If compressed to negative $x$, the force pushes toward equilibrium.

Deriving the equation of motion

Apply Newton's second law:

$F=ma$

For the spring force,

-kx=mrac{d^2x}{dt^2}

Rearrange:

rac{d^2x}{dt^2}+rac{k}{m}x=0

This has the SHM form

rac{d^2x}{dt^2}+omega^2x=0

so

omega=sqrt{rac{k}{m}}

The angular frequency depends on the stiffness of the spring and the mass attached to it. A stiffer spring gives faster oscillation. A larger mass gives slower oscillation.

Period of a mass-spring oscillator

The period is

T=rac{2pi}{omega}

Substitute $omega=sqrt{k/m}$:

T=2pisqrt{rac{m}{k}}

This equation says the period increases with mass and decreases with spring stiffness. Doubling the mass does not double the period; the dependence is through a square root.

Horizontal and vertical springs

For a horizontal frictionless spring, equilibrium is usually the natural length of the spring. For a vertical spring, gravity stretches the spring to a new equilibrium position. The equilibrium stretch satisfies

$kx_0=mg$

If displacement is measured from this new equilibrium, the motion is still SHM with the same angular frequency:

omega=sqrt{rac{k}{m}}

Gravity shifts the equilibrium but does not change the oscillation frequency in the ideal linear model.

Solution and phase

The displacement can be written

$x(t)=Acos(omega t+phi)$

The velocity is

$v(t)=-Aomegasin(omega t+phi)$

The constants $A$ and $phi$ depend on how the mass is released. If the mass is pulled to $x=A$ and released from rest, then $phi=0$ is convenient. If the mass passes through equilibrium at $t=0$ with maximum positive velocity, a sine form may be simpler.

Energy in the spring system

The spring potential energy is

U_s=rac{1}{2}kx^2

The kinetic energy is

K=rac{1}{2}mv^2

In the ideal system, total mechanical energy is conserved:

E=rac{1}{2}mv^2+rac{1}{2}kx^2

At maximum displacement, $v=0$ and

E=rac{1}{2}kA^2

At equilibrium, $x=0$ and speed is maximum:

E=rac{1}{2}mv_{max}^2

Thus

v_{max}=Asqrt{rac{k}{m}}=Aomega

Physical interpretation

The mass does not move fastest at the endpoints. It stops there momentarily before reversing direction. It moves fastest at equilibrium, where all the energy is kinetic. The acceleration is greatest at the endpoints because the spring force has greatest magnitude there.

The big idea

The mass-spring oscillator is the cleanest physical example of simple harmonic motion. Hooke's law supplies a linear restoring force, Newton's second law produces the SHM differential equation, and energy conservation reveals how kinetic and spring potential energy trade back and forth throughout the cycle.