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12.4More about energy in SHM

In SHM we know that the displacement is given by x = x 0 sin(ω t + ϕ) and the velocity by v = ω x 0 cos(ω t + ϕ). The kinetic energy is then given by

E K=1 2 m v 2=1 2 m ω 2 x 0 2 cos 2⁡(ω t+ϕ)

This can be rewritten as

E K=1 2 m ω 2 x 0 2(1−sin 2⁡(ω t+ϕ))

E K=1 2 m ω 2 x 0 2−1 2 m ω 2 x 2.

The maximum kinetic energy is then 1 2 m ω 2 x 0 2, and so this must also be the total energy of the system:

E T=1 2 m ω 2 x 0 2

The total energy is the sum of the potential and kinetic energies:

E T=E P+E K

Therefore

E p=1 2 m ω 2 x 0 2−(1 2 m ω 2 x 0 2−1 2 m ω 2 x 2)

E p=1 2 m ω 2 x 2

KEY POINT

In SHM the energies are E p=1 2 m ω 2 x 2, E K=1 2 m ω 2(x 0 2−x 2) and E T=1 2 m ω 2 x 0 2.

Since E K=1 2 m v 2, we find a very useful formula for velocity in terms of displacement:

v=±ω x 0 2−x 2

The plus or minus sign is needed because for any given displacement x, the particle may be going one way or the opposite way. This equation allows us to see that at the extremes of the motion, x = ±x 0, and so v = 0 as we expect. This formula allows for a much simpler solution to a problem such as problem 24 in the previous chapter. Using the data for that problem the velocity is simply

v=±ω x 0 2−x 2=±π 2 0.12 2−0.06 2=±0.16 m s−1.

Worked Example 12.7

The graph of Figure 12.37 shows the variation with the square of the displacement (x 2) of the potential energy E P of a particle of mass 40 g that is executing SHM. Using the graph, determine:

athe period of oscillation

bthe maximum speed of the particle during an oscillation.

Figure 12.37: For worked example 12.7.

Answer

aThe maximum potential energy is E P=1 2 m ω 2 x 0 2.

From the graph, the maximum potential energy is 80 mJ (= 8.0 × 10−2 J) and the amplitude is 4.0 cm 2=2.0 cm(=2.0×10−2 m).

The mass of the particle is 0.040 kg, so:

80×10−3=1 2×0.040×ω 2×x 0 2

Rearranging:

ω 2=2×80×10−3 0.040×(2.00×10−2)2 ω 2=10 4 s−2 ω=10 2 rad s−1

The period is T=2 π ω, so:

T=2 π 10 2=0.063 s

bThe maximum speed is found from:

v max=ω x 0=100×2.00×10−2=2.00 m s−1

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Worked Example 12.8

In SHM the maximum speed is 4.0 m s−1 and the maximum acceleration is 35 m s−2. Determine:

athe period of the motion

bthe amplitude of oscillations.

Answer

a v max=ω x 0 and a max=ω 2 x 0

So, a max v max=ω 2 x 0 ω x 0=ω

Hence, ω=35 4.0=8.75 s−1

The period is T=2 π ω=2 π 8.75≈0.72 s.

bFrom v max=ω x 0 we find 4.0=8.75×x 0

So, x 0=4.0 8.75=0.46 m.

Worked Example 12.9

The graph of Figure 12.38 shows the variation with time t of the kinetic energy E K of a particle of mass 0.25 kg that is undergoing SHM.

Figure 12.38: For worked example 12.9.

For this motion, determine:

athe period

bthe amplitude

cthe kinetic energy when the displacement is 0.080 m.

Answer

aAt t = 0 the kinetic energy is zero, meaning that the particle is at one extreme of the oscillation. It is zero again at about 0.175 s when it is at the other end. To find the period the particle has to return to its original position and that happens at 0.35 s, so T = 0.35 s.

bThe angular frequency is found from ω=2 π T:

ω=2 π 0.35=17.95≈18 rad s−1

The maximum kinetic energy is given by E K max=1 2 m ω 2 x 0 2 and so:

0.80=1 2×0.25×17.95 2×x 0 2

Rearranging:

x 0=2×0.80 0.25×17.95 2=0.141≈0.14 m

cWe know the displacement so use a formula that gives kinetic energy in terms of displacement:

E K=1 2 m ω 2(x 0 2−x 2)

This gives:

E K=1 2×0.25×17.95 2(0.141 2−0.080 2)=0.54 J

Exam tip

It is important to understand how to find the period from this graph.

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NATURE OF SCIENCE

The complex can be understood in terms of the simple.

The equation for SHM can be solved in terms of simple sine and cosine functions. These simple solutions help physicists to visualise how an oscillator behaves. Although real oscillations are very complex, a powerful mathematical machinery called Fourier analysis allows the decomposition of complex oscillations, sounds, noise and waves in general, in terms of sines and cosines. Energy exchange in oscillating electrical circuits is modelled using this type of analysis. Therefore the simple descriptions used in this topic can also be used in more complex problems as well.

TEST YOUR UNDERSTANDING

34Figure 12.39 shows the variation with displacement x of the acceleration a of a body of mass 0.150 kg.

Figure 12.39: For question 34.

aUse the graph to explain why the motion of the body is SHM.

Determine the following:

bthe period of the motion

cthe maximum velocity of the body during an oscillation

dthe maximum net force exerted on the body

ethe total energy of the body.

35Figure 12.40 shows the variation with time of the kinetic energy of a particle that is undergoing simple harmonic oscillations.

Figure 12.40: For question 35.

aUse the graph to determine the period of oscillation.

bDraw a graph to show the variation with time of the potential energy.

cWhat is the potential energy at

i1.0 s

ii2.0 s?

36In a SHM the amplitude of oscillations is 15 cm and the period is 2.0 s. The mass of the particle is 0.12 kg.

aWhat is the potential energy when the displacement is 9.0 cm?

bWhat is the kinetic energy when the displacement is 6.0 cm?

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37The total energy of a body executing SHM is 18 J. The amplitude of oscillations is 12 cm. What is the potential energy of the body when a chevron is needed by 37 displacement is 8.0 cm?

38The total energy of a body executing SHM is 12 J. The amplitude of oscillations is 8.0 cm. What is the kinetic energy of the body when the displacement is 4.0 cm?

39The total energy of a particle in SHM is 8.0 J. The amplitude of the motion is 5.0 cm. What is the displacement when the kinetic energy is 6.0 J?

40A point undergoes SHM oscillations with period 1.5 s. The speed when the displacement is 4.0 cm is 18 cms−1. Determine the amplitude of the oscillations.

41At a particular displacement x in a SHM, the potential energy equals the kinetic energy. The amplitude of the motion is 4.2 cm. What is x?