When a weight is added to a spring and stretched, the released spring will follow a back and forth motion.

At the equilibrium position, velocity is at its maximum. At maximum displacement, spring force and acceleration reach a maximum.

It is found that the force applied is directly proportional to the distance the spring stretches (or is compressed).

As long as force remains proportional to distance, a plot of force vs. distance is a straight line. Actually there is a point at which force vs. distance are no longer proportional, this is called the “proportionality limit”.

Once the force is removed, the object will return to its original shape. If the “elastic limit” is exceeded, the object will not return to its original shape and will be permanently deformed.

For most springs: F = kx x is how much the spring is displaced from its original length, k is the spring constant (N/m), F is the force. This is known as Hooke’s Law.

A spring that behaves according to Hooke’s Law is called an ideal spring.

If a mass of 0.55 kg attached to a vertical spring stretches the spring 2.0 cm from its original equilibrium position, what is the spring constant?

F= kx 0.55 kg x 10 = k(0.02) 5.5 = k(0.02) k = 275 N/m

The spring constant k is often referred to as the “stiffness” of the spring.

A force must be applied to a spring to stretch or compress it. By Newton’s third law, the spring must apply an equal force to whatever is applying the force to the spring.

This reaction force is often called the “restoring force” and is represented by the equation F = -kx.

This force varies with the displacement. Therefore the acceleration varies with the displacement.

When the restoring force has the mathematical form given by F = -kx, the type of motion resulting is called “simple harmonic motion”.

A graph of this motion is sinusoidal. When an object is hung from a spring, the equilibrium position is determined by how far the weight stretches the spring initially.

When a spring is stretched or compressed it has elastic potential energy.

PE elastic = 1/2 kx 2 where k is the spring constant, and x is the distance the spring is compressed or stretched beyond its unstrained length. The unit is the joule (J).

When external nonconservative forces do no net work on a system then total mechanical energy must be conserved. E f = E 0

Total mechanical energy = translational kinetic energy + rotational kinetic energy + gravitational potential energy + elastic potential energy.

If there is no rotation, this becomes this equation: E = 1/2 mv 2 + mgh + 1/2 kx 2

A 0.20-kg ball is attached to a vertical spring. The spring constant is 28 N/m. The ball, supported initially so that the spring is neither stretched nor compressed, is released from rest. How far does the ball fall before being momentarily stopped by the spring?

1/2 mv 2 + mgh + 1/2 kx 2 = 1/2 mv 2 + mgh + 1/2 kx 2

1/2 mv 2 + mgh + 1/2 kx 2 = 1/2 mv 2 + mgh + 1/2 kx 2 mgh = 1/2 kx 2 (0.2)10h = 1/2 (28)x 2 2h = 14x 2 1 = 7x 1/7 = x

A simple pendulum is a mass m suspended by a pivot P. When the object is pulled to one side and released, it will swing back and forth in a motion approximating simple harmonic motion.

When a pendulum swings through small angles, 2πf = √g/L. f is frequency, g is 9.80, and L is length.

2πf = √g/L Mass is algebraically eliminated, and it has no bearing on the frequency of a pendulum.

In some instances it is more useful to use the period T of vibration rather than frequency f. T = 2π √L/g

Determine the length of a simple pendulum that will swing back and forth in simple harmonic motion with a period of 1.00 s.

A pendulum attached to the ceiling almost touches the floor and its period is 12 s. How high is the ceiling?

A pendulum can be a real object, in which case it is called a physical pendulum.

In reality, an object in simple harmonic motion will not vibrate forever. Friction, or some such force, will decrease the velocity and amplitude of the motion. This is called damped harmonic motion.

Formulas for frequency and period of an oscillating spring: 2πf = √ k/m T = 2π√ m/k

The body of a 1275 kg car is supported on a frame by four springs. Two people riding in the car have a combined mass of 153 kg. When driven over a pothole, the frame approximates simple harmonic motion with a period of s. Find the spring constant of a single spring.