Energy in SHM

Kinetic and potential energy

Narrated video. Transcript:

A block oscillates on a spring on a smooth floor. As it moves, its speed keeps changing, and so does the stretch of the spring. Let us follow the energy.

The kinetic energy is half m v squared. With v equals minus ωA sin of ωt plus φ, this is half m ω squared A squared sin squared of the phase. Since mω squared is k, K equals half k A squared sin squared of ωt plus φ.

The spring force, minus k x, is a conservative force, so the system has potential energy. U equals half k x squared, which is half k A squared cos squared of ωt plus φ.

Now add them. Sin squared plus cos squared is always one, so the total energy E equals half k A squared. It does not change with time.

Watch the energy bars. At the ends the block stops, and all the energy is potential. At the centre the spring is relaxed, and all of it is kinetic. The total bar never moves.

Plot K and U against time. Both swing between zero and E, but each repeats twice in every oscillation. Their period is T over 2.

Now plot them against displacement x. U is a parabola, half k x squared, opening upwards. K is an upside-down parabola, E minus U. Their sum is a flat line at E.

The two parabolas cross where K equals U, each half of E. That happens where half k x squared equals a quarter k A squared: at x equals A over root 2, about 0.71 A, not at A over 2.

Remember: the total energy depends on the square of the amplitude. Double the amplitude, and the energy of the oscillation becomes four times bigger.

Worked example

Narrated video. Transcript:

A 1 kg block is attached to a spring of spring constant 50 newtons per metre, on a frictionless floor. At t equals zero it is pulled 10 cm from equilibrium and released.

Find its kinetic, potential and total energy when it is 5 cm from the mean position.

The amplitude is 10 cm, because that is where the block was released from rest. So the total energy is half k A squared: half times 50 times 0.1 squared, which is 0.25 joules.

At x equals 5 cm, the potential energy is half times 50 times 0.05 squared, which is 0.0625 joules.

The total stays fixed, so the kinetic energy is what is left: 0.25 minus 0.0625, which is 0.1875 joules.

Check with the speed formula. ω is the square root of 50, about 7.07 radians per second. v equals ω times the square root of A squared minus x squared, about 0.61 metres per second. Half times 1 times 0.61 squared gives 0.19 joules. It matches.

Notice: at half the amplitude, the spring holds only a quarter of the energy. Three quarters is still kinetic. Energy goes as x squared, not as x.

On a real floor, friction slowly turns this energy into heat, and the amplitude shrinks. Here we keep the floor smooth, so E stays at half k A squared for ever.

Try it: energy against x

Interactive simulation. Things to try:

• Press Release. The white dot rides on U = ½kx², the amber dot on K = E − U.
• The green line is the total energy E = ½kA². It never moves while the block oscillates.
• Find where the two curves cross. Is it at A/2?
• Double A and read the new E in the corner of the graph.

Try it: energy bars

Interactive simulation. Things to try:

• This is the worked example: k = 50 N/m, m = 1 kg, A = 0.1 m. Total energy 0.25 J.
• Press Release and watch K and V swap, while K + V stays fixed.
• Change the mass. Does the total energy change? What about the maximum speed?

Half and half

A particle performs SHM with amplitude A. At what displacement are its kinetic and potential energies equal?

1. x = ±A/4
2. x = ±A/2
3. x = ±A/√2
4. x = ±A

Maximum speed from energy

A 0.2 kg mass on a spring of spring constant 80 N/m oscillates on a smooth surface with amplitude 5 cm. What is its maximum speed?

1. 0.5 m/s
2. 1 m/s
3. 2 m/s
4. 10 m/s

Summary

• Kinetic energy: K = ½mv² = ½kA² sin²(ωt + φ)
• Potential energy: U = ½kx² = ½kA² cos²(ωt + φ)
• Total energy: E = K + U = ½kA²
• K = U at: x = ±A/√2

• In SHM, energy keeps changing between kinetic (largest at the mean position) and potential (largest at the extremes).
• With no friction, the total energy is constant: E = ½kA², proportional to the square of the amplitude.
• K and U each vary with a period of T/2, half the period of the motion.
• Against x, U is a parabola, K is an inverted parabola, and their sum is a flat line.