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12.4More about energy in SHM
In SHM we know that the displacement is given by x = x0 sin(Ït + Ï) and the velocity by v = Ïx0 cos(Ït + Ï). The kinetic energy is then given by
EK=12mv2=12mÏ2x20cos2(Ït+Ï)
This can be rewritten as
EK=12mÏ2x20(1âsin2(Ït+Ï))
EK=12mÏ2x20â12mÏ2x2.
The maximum kinetic energy is then 12mÏ2x20, and so this must also be the total energy of the system:
ET=12mÏ2x20
The total energy is the sum of the potential and kinetic energies:
ET=EP+EK
Therefore
Ep=12mÏ2x20â(12mÏ2x20â12mÏ2x2)
Ep=12mÏ2x2
KEY POINT
In SHM the energies are Ep=12mÏ2x2, EK=12mÏ2(x20âx2) and ET=12mÏ2x20.
Since EK=12mv2, we find a very useful formula for velocity in terms of displacement:
v=Â±Ïx20âx2âââââââ
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 = Â±x0, 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=Â±Ïx20âx2âââââââ=Â±Ï20.122â0.062ââââââââââââ=Â±0.16Â m sâ1.
WORKED EXAMPLE 12.7
The graph of Figure 12.37 shows the variation with the square of the displacement (x2) of the potential energy EP 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 EP=12mÏ2x02.
From the graph, the maximum potential energy is 80 mJ (= 8.0 Ã 10â2 J) and the amplitude is 4.0Â cm2ââââââââ=2.0Â cm(=2.0Ã10â2Â m).
The mass of the particle is 0.040 kg, so:
80Ã10â3=12Ã0.040ÃÏ2Ãx20
Rearranging:
Ï2Ï2Ï=2Ã80Ã10â30.040Ã(2.00Ã10â2)2=104Â sâ2=102Â rad sâ1
The period is T=2ÏÏ, so:
T=2Ï102=0.063Â s
bThe maximum speed is found from:
vmax=Ïx0=100Ã2.00Ã10â2=2.00Â m sâ1
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WORKED EXAMPLE 12.8
In SHM the maximum speed is 4.0 m sâ1 and the maximum acceleration is 35 m sâ2. Determine:
athe period of the motion
bthe amplitude of oscillations.
Answer
avmax=Ïx0Â andÂ amax=Ï2x0
So, amaxvmax=Ï2x0Ïx0=Ï
Hence, Ï=354.0=8.75Â sâ1
The period is T=2ÏÏ=2Ï8.75â0.72Â s.
bFrom vmax=Ïx0 we find 4.0=8.75Ãx0
So, x0=4.08.75=0.46Â m.
WORKED EXAMPLE 12.9
The graph of Figure 12.38 shows the variation with time t of the kinetic energy EK 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 Ï=2ÏT:
Ï=2Ï0.35=17.95â18Â rad sâ1
The maximum kinetic energy is given by EKmax=12mÏ2x20 and so:
0.80=12Ã0.25Ã17.952Ãx20
Rearranging:
x0=2Ã0.800.25Ã17.952ââââââââââââ=0.141â0.14Â m
cWe know the displacement so use a formula that gives kinetic energy in terms of displacement:
EK=12mÏ2(x20âx2)
This gives:
EK=12Ã0.25Ã17.952(0.1412â0.0802)=0.54Â 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 cmsâ1. 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?
