Energy in simple harmonic motion
The kinetic energy of any body of
mass m is given by kinetic energy = ½ mv
2So, in simple harmonic motion:
Kinetic energy = ½ mω2(r2 - x2)
Now the
maximum value of the kinetic energy will occur when x = 0, and this will be equal to the
total energy of the body. Therefore:
Total energy = ½ mω2r2
Therefore, since:
potential energy = total energy – kinetic energy, the potential energy at any point will be
given by:
Potential energy = ½ mω2x2
Graphs
of the variation of potential energy, kinetic energy and the total energy are shown in the
accompanying diagram (Figure 1).
Example problems
A 2 kg body panel of a car oscillates with a frequency of 2 Hz and amplitude 2.5 cm. If the oscillations are assumed to be simple harmonic motion and undamped, calculate
(a) the maximum velocity of the panel,
(b) the total energy of the panel
(c) the maximum potential energy of the panel
(d) the kinetic energy of the panel 1.0 cm from its equilibrium position.
(a) From v = +ω(r
2 - x
2)
1/2 the maximum velocity occurs when x = 0, at the centre of the oscillation.
Therefore maximum velocity = +12.6 x 2.5 = 31.5 cms
-1 = 0.315 ms
-1
(b) Total energy = ½ mω
2 = ½ x 2 x 12.6 x 12.6 x 0.025 x 0.025 = 0.1 J
(c) Maximum p.e = total energy = 0.1 J
(d) Kinetic energy = ½ mω
2(r
2 - x
2) = ½x2x12.6x12.6(0.025
2 — 0.01
2) = 0.0833 J
Note the conversion of centimetres to
metres for the energy calculations.
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