AP Physics 1 · Lesson 3

Graphs and energy of oscillators

Know Know it

If a cart on a spring is released from rest at x = +A at t = 0, its position follows a cosine curve, x=Acos⁡(2πt/T)x = A\cos(2\pi t/T). It passes x = 0 at T/4, reaches −A at T/2, and is back at +A at T.

The velocity is the slope of the position–time graph: zero at each peak and trough, largest in size where the graph crosses x = 0. The acceleration graph is the position graph flipped upside down and scaled, because a=−kmxa = -\frac{k}{m}x.

With no friction the total mechanical energy is fixed at E=12kA2E = \frac{1}{2}kA^2, and at every position 12kx2+12mv2=12kA2\frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \frac{1}{2}kA^2. At the turning points the energy is all spring potential energy; at equilibrium it is all kinetic, so vmax=Ak/mv_{max} = A\sqrt{k/m}. For a pendulum, gravitational energy mghmgh plays the part of the spring energy.

Spring potential energy U against position x for k = 50 N/m and an amplitude of 0.20 m−0.3−0.2−0.10.10.20.30.20.40.60.811.20x (m)U (J)(−0.20, 1.0)(0.20, 1.0)
Spring potential energy U against position x for k = 50 N/m and an amplitude of 0.20 m An upward-opening parabola with its lowest point at (0, 0), rising to 1.0 J at x = −0.20 m and at x = +0.20 m. A dashed horizontal line at 1.0 J marks the total energy; at any x, the gap between that line and the curve is the kinetic energy.

Worked example Easy

A 0.40 kg cart on a level, frictionless track is attached to a spring with k = 120 N/m. It oscillates with an amplitude of 0.15 m. What is the total mechanical energy of the cart–spring system?

  1. 9.00 J
  2. 2.70 J
  3. 18.0 J
  4. 1.35 JAnswer

How to solve it

  1. At a turning point all the energy is spring energy: E=12kA2E = \frac{1}{2}kA^2.
  2. E=12(120)(0.0225)=1.35E = \frac{1}{2}(120)(0.0225) = 1.35 J. The mass would matter only for a speed.

Why each choice is right or wrong

  • A. Incorrect. 9.00 J is one-half k times A without squaring the amplitude, so its unit is not even joules.
  • B. Incorrect. 2.70 J leaves out the factor of one-half in the spring energy.
  • C. Incorrect. 18.0 J is k times A, the force at the turning point in newtons, not an energy.
  • D. Correct. E=12kA2=12(120)(0.15)2=1.35E = \frac{1}{2}kA^2 = \frac{1}{2}(120)(0.15)^2 = 1.35 J; the cart’s mass does not enter.

Trap Halfway out is not half the energy

At x = A/2 the spring holds 12k(A/2)2\frac{1}{2}k(A/2)^2, only one quarter of the total energy, so three quarters is still kinetic. Energy goes as the square of the displacement, so halfway in position is not halfway in energy.

Explain Explain it

Task verbs: describe, explain

Doubling the amplitude of the same oscillator doubles the maximum speed, multiplies the total energy by four and leaves the period unchanged. To find the speed at any position, subtract the spring energy there from the total: 12mv2=12k(A2−x2)\frac{1}{2}mv^2 = \frac{1}{2}k(A^2 - x^2).

Friction or air drag slowly removes mechanical energy; this is damping. The amplitude shrinks, and because E∝A2E \propto A^2, the fraction of energy lost is larger than the fraction of amplitude lost.

Worked example Medium

A 0.50 kg block on a frictionless horizontal surface oscillates on a spring with k = 72 N/m and an amplitude of 0.10 m. What is the block’s speed when it is 0.06 m from equilibrium?

  1. 1.2 m/s
  2. 0.48 m/s
  3. 0.96 m/sAnswer
  4. 0.72 m/s

How to solve it

  1. Energy: 12mv2+12kx2=12kA2\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2, so v2=km(A2−x2)v^2 = \frac{k}{m}(A^2 - x^2).
  2. km=72/0.50=144\frac{k}{m} = 72/0.50 = 144 per s², and A2−x2=0.0100−0.0036=0.0064A^2 - x^2 = 0.0100 - 0.0036 = 0.0064 m².
  3. v=(144)(0.0064)=0.9216=0.96v = \sqrt{(144)(0.0064)} = \sqrt{0.9216} = 0.96 m/s.

Why each choice is right or wrong

  • A. Incorrect. 1.2 m/s is the maximum speed, which the block has only at equilibrium, not 0.06 m away from it.
  • B. Incorrect. 0.48 m/s multiplies k/m=12\sqrt{k/m} = 12 by A − x = 0.04 m; the energy depends on the squares, A2−x2A^2 - x^2, not on A − x.
  • C. Correct. v=km(A2−x2)=144(0.0100−0.0036)=0.96v = \sqrt{\frac{k}{m}(A^2 - x^2)} = \sqrt{144(0.0100 - 0.0036)} = 0.96 m/s.
  • D. Incorrect. 0.72 m/s multiplies k/m\sqrt{k/m} by x; the speed depends on how far the block is from the turning point, through A2−x2A^2 - x^2.

Describe Your turn

A block on a frictionless horizontal spring is released from rest at x = +A. Describe how the block’s kinetic energy and the spring’s potential energy change as the block moves from x = +A to x = 0, and explain why their sum stays the same.

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Compare with a model answer

A full answer does this

  • Says the kinetic energy rises from zero to its greatest value
  • Says the spring potential energy falls from its greatest value to zero
  • Says the sum stays the same because nothing takes energy out of or puts energy into the block–spring system

Model answer

At x = +A the block is at rest, so its kinetic energy is zero and the spring’s potential energy is at its greatest. As the block moves toward x = 0 the spring relaxes: its potential energy falls to zero at x = 0, while the block speeds up and its kinetic energy rises to its greatest value there. The sum stays the same because the surface is frictionless and no outside force does work on the block–spring system, so energy only changes form from spring energy to kinetic energy.

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