9478 / 2027
Oscillations overview

Lesson 4 of 6

Follow the energy exchange

  • I can explain kinetic and potential energy interchange during SHM.Syllabus

    Syllabus 9478, 9(h). Explain kinetic and potential energy interchange during SHM.

An ideal free oscillator repeatedly exchanges kinetic and potential energy. Their total stays constant when no energy is supplied or dissipated.

Choose the horizontal spring and cart as the system. Take elastic potential energy at equilibrium as zero. The cart stays at one height, so gravitational energy is unchanged. Use a linear spring, negligible losses and no continuing drive.

Obtain the potential energy from work

The work needed to stretch or compress a linear spring from equilibrium is the area under its force-extension graph. This gives:

Ep = ½kx2 = ½mω2x2
Ek = ½mv2

The second form uses k = mω2 for this spring-cart model. Squaring displacement makes Ep non-negative on either side of equilibrium. At either extreme, velocity is zero, so the total energy is:

Etotal = ½mω2x02

This is derived from the spring work and the turning-point condition. It is not an additional energy to add to Ek + Ep.

Worked energy exchange

Track the same 8.00 mJ around the motion

For m = 0.400 kg, ω = 4.00 rad/s and x0 = 0.0500 m:

Etotal = ½(0.400)(4.00)2(0.0500)2
= 0.00800 J = 8.00 mJ

Here 1 mJ = 0.001 J. At equilibrium, Ep = 0 and all 8.00 mJ is kinetic in this reference. At either extreme, Ek = 0 and all 8.00 mJ is elastic potential.

Energy changes form as the signed displacement changes

For this horizontal spring-cart, take the spring's potential energy at equilibrium as zero. Gravitational energy is unchanged and dissipation is negligible. 1 mJ = 0.001 J.

Ep: solid, elastic potential

Ek: long dash, kinetic

Total: dotted, 8.00 mJ

Nonnegative kinetic and potential energies add to eight millijoulesEnergy in millijoules is plotted against signed displacement from negative 0.0500 to positive 0.0500 metres. The solid elastic potential-energy parabola rises from zero at equilibrium to eight millijoules at both extremes. The long-dashed kinetic-energy parabola falls from eight millijoules at equilibrium to zero at both extremes. Their sum is the dotted horizontal eight-millijoule line. The curves use the same energy scale and are drawn only over the physically allowed displacement interval; neither is negative. At displacement positive 0.0300 metres their values are 2.88 and 5.12 millijoules. Horizontal position is displacement, not time, and the account is the same on either direction of travel.-0.050-0.0250+0.025+0.05002468Energy / mJDisplacement x / m

At x = +0.0300 m, 2.88 mJ potential + 5.12 mJ kinetic = 8.00 mJ. At either extreme, the cart is momentarily at rest and the energy is entirely potential in this reference.

Both energies are plotted against signed displacement, with a horizontal constant-total line. The allowed motion lies between the two turning points. These are energy-position curves, not displacement-time graphs.

At x = +0.0300 m, the speed magnitude is 0.160 m/s:

Ep = ½(6.40)(0.0300)2
= 0.00288 J = 2.88 mJ
Ek = ½(0.400)(0.160)2
= 0.00512 J = 5.12 mJ
Ek + Ep = 8.00 mJ

The same energy values occur on the outward and return passes. Velocity changes sign, but kinetic energy depends on v2.

Try it step by step

Work it out in steps: Total energy 1/2 m omega2 x02 in J, then Kinetic energy at x = x0 / 2, which is 3/4 of the total, in J.

  1. Problem 1. A 0.20 kg mass oscillates with omega = 10 rad/s and amplitude 5.0 cm. Find the total energy and the kinetic energy when the displacement is half the amplitude.
  2. Problem 2. A 0.50 kg mass oscillates with omega = 4.0 rad/s and amplitude 10 cm. Find the total energy and the kinetic energy at half the amplitude.
  3. Problem 3. A 0.10 kg mass oscillates with omega = 20 rad/s and amplitude 3.0 cm. Find the total energy and the kinetic energy at half the amplitude.
Show the answer

Problem 1.

Total energy 1/2 m omega2 x02 in J: E = 1/2 x 0.20 x 102 x 0.0502 = 0.025 J

Kinetic energy at x = x0 / 2, which is 3/4 of the total, in J: KE = 1/2 m omega2 (x02 - x2) = 3/4 x 0.025 = 0.019 J (2 sf)

Problem 2.

Total energy 1/2 m omega2 x02 in J: E = 1/2 x 0.50 x 4.02 x 0.102 = 0.040 J

Kinetic energy at x = x0 / 2, which is 3/4 of the total, in J: KE = 3/4 x 0.040 = 0.030 J (2 sf)

Problem 3.

Total energy 1/2 m omega2 x02 in J: E = 1/2 x 0.10 x 202 x 0.0302 = 0.018 J

Kinetic energy at x = x0 / 2, which is 3/4 of the total, in J: KE = 3/4 x 0.018 = 0.0135 J, which is 0.014 J to 2 sf

Over the allowed displacement range, Ep is an upward parabola and Ek is the total minus that parabola. Both remain non-negative. The energy pattern repeats after T/2 because the squares x2 and v2 repeat then, even though the full signed motion takes T.

Change amplitude within the same linear model

Doubling amplitude to 0.100 m with the same mass and stiffness doubles maximum speed from 0.200 to 0.400 m/s and maximum acceleration magnitude from 0.800 to 1.60 m/s2.

Etotal is proportional to x02
New energy = 4 × 8.00 mJ = 32.0 mJ = 0.0320 J

The period stays the same only while the same linear restoring model remains valid. A larger displacement that changes the restoring relationship need not preserve that result.

For a vertical oscillator, the potential-energy account must include both spring and gravitational contributions about the shifted equilibrium. Do not label all of that combined change as spring energy alone.

Check your understanding

A linear undamped oscillator has total mechanical energy 8.00 mJ at amplitude 0.0500 m. Its amplitude doubles while mass and stiffness stay unchanged and the linear model remains valid. What changes?

  1. Energy becomes 16.0 mJ and period doubles.
  2. Energy becomes 32.0 mJ and the period doubles with the amplitude.
  3. Energy becomes 32.0 mJ while the period stays the same.
Show the answer

Energy becomes 32.0 mJ while the period stays the same.

Doubling amplitude gives four times the energy. With unchanged mass and linear stiffness, angular frequency and period stay unchanged within the stated model.