Restoring Force

Examples: Springs and Pendulums

When you grab an object attached to a spring, pull it a distance from its equilibrium position, and release it, the restoring force will pull the object back to equilibrium. For a spring-mass system in a horizontal table, the only force acting on the mass in the direction of displacement is the restoring force exerted by the spring.

Using Newton's Second Law we can set up an equation for the motion of the object. It is important to note that in this article we will not be using the vector form of the second law as in this case we are only studying the magnitude of the restoring force in one dimension. The direction of the restoring force will always be antiparallel to the displacement of the object. This can be proved experimentally and is a feature of Hooke's Law. The restoring force acting on the spring-mass system depends on the spring constant and the object's displacement from the equilibrium position.

The expression for the force is

$$F_x=ma_x.$$

Substituting the spring force for \(F_x\) and the \(a_x\) for the second derivative with respect to time we obtain

$$-kx=m\frac{\operatorname d^2x}{\operatorname dt^2}.$$

Rearranging for the second derivative then yields the equation

$$\frac{\operatorname d^2x}{\operatorname dt^2}=-\frac kmx,$$

where \(m\) is the mass of the object at the end of the spring in kilograms \((\mathrm{kg})\), \(a_x\) is the acceleration of the object on the \(\text{x-axis}\) in meters per second squared \((\frac{\mathrm m}{\mathrm s^2})\), \(k\) is the spring constant that measures the stiffness of the spring in newtons per meter \((\frac{\mathrm N}{\mathrm m})\), and \(x\) is the displacement in meters \((\mathrm m)\).

The above expression looks a lot like the differential equation for simple harmonic motion, so the spring-mass system is a harmonic oscillator, where its angular frequency can be expressed in the following equation:

$$\omega^2=\frac km,$$ or explicitly as

$$\omega=\sqrt{\frac km}.$$

The second example we will be discussing is the case of a simple pendulum. A simple pendulum consists of a mass that swings about an equilibrium position while hanging from a rod. The restoring force is exerted by gravity.

As we can see in the above image, the restoring force is the component of the force of gravity that is antiparallel to the displacement of the pendulum. This equation is obtained using trigonometric relationships and the geometry of the system.

$$\sin\left(\theta\right)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}=\frac{F_{antiparallel}}{-mg}$$

$$F_{antiparallel}=F_{restoring}=-mg\sin\left(\theta\right)$$

Where \(m\) is the mass of the pendulum in kilograms, \((\mathrm{kg})\), \(\mathrm g\) is the acceleration due to gravity in meters per second squared, \((\frac{\mathrm m}{\mathrm s^2})\), and \(\theta\) is the angle between the equilibrium position and the displacement position in degrees or radians, \((^\circ\;\mathrm{or}\;\mathrm{rad})\).

For an object to be considered a harmonic oscillator, the restoring force must be proportional to the displacement. In this case, it is proportional to the force of gravity and the sine of the displacement angle \(\theta\). However, there are some cases where the motion of a simple pendulum is considered to be moving in simple harmonic motion. When the displacement angle is very small, the sine of the displacement angle can be approximated to be the angle itself, such that \(\sin\left(\theta\right)\approx\theta\). In this case, the restoring force is proportional to the displacement.

Now we can examine the differential equation used to describe the motion of a simple pendulum for a small displacement angle. First, we need to introduce the concept of arc length in order to solve the differential equation.

Again, we begin the approach with Newton's Second Law, which is given by

$$F=ma.$$ Substituting the restoring force and the second derivative of displacement with respect to time we obtain

$$-mg\sin\left(\theta\right)=m\frac{\operatorname d^2s}{dt^2},$$ where \[\sin\left(\theta\right)\approx\theta,\]

\[\frac{\operatorname d^2s}{dt^2}=-g\theta,\] and where \(s=L\theta.\)

The length of the pendulum is constant so only the displacement angle will change with time. The equation then becomes

$$L\frac{\operatorname d^2\theta}{dt^2}=-g\theta.$$

Rearranging for the acceleration we obtain

$$\frac{\operatorname d^2\theta}{dt^2}=-\frac gL\theta.$$

The above expression looks a lot like the differential equation for simple harmonic motion, so the single pendulum with a small displacement angle is a harmonic oscillator, where its angular frequency is expressed as

$$\omega^2=\frac gL,$$

or explicitly as

$$\omega=\sqrt{\frac gL}.$$