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12.2Details of simple harmonic motion
This section is a quantitative discussion of SHM. We saw that SHM takes place whenever the acceleration is opposite to and proportional to the displacement. We can express this mathematically as
a=âÏ2x
We learned in the previous chapter that the angular frequency Ï is related to the period T by Ï=2ÏT.
What is the equation that gives the displacement as a function of time? Using calculus, it can be shown that the defining equation a = âÏ2 x implies that the displacement x is given by the formula
x=x0sin(Ït+Ï)
where Ï is called the phase angle and x0 is the amplitude.
The phase angle has to do with the position and velocity of the particle at zero time. In other words, it has to do with when you start the stopwatch as you observe the oscillations. So, the same oscillation may be described by x1 = x0 sin(Ït + Ï1) or x2 = x0 sin(Ït + Ï2).
The phase difference ÎÏ we defined in SectionÂ 12.1 is simply ÎÏ = |Ï1 â Ï2|.
From x = x0 sin(Ït + Ï) and again using calculus, we can find the velocity and the acceleration. The results are:
v=Ïx0cos(Ït+Ï)
a=âÏ2x0sin(Ït+Ï)
Notice that the last equation says a = âÏ2x as it should be. These equations imply that the maximum values of velocity and acceleration are:
vmax=Ïx0
amax=Ï2x0
CHECK YOURSELF 6
In a SHM the amplitude of oscillations is 0.20 m and the period is 1.5 s.
aWhat is the initial displacement if the phase angle is:
izero
iiÏ2?
bWrite down the equation for displacement in the case of a phase angle of Ï2 in its simplest form.
CHECK YOURSELF 7
In a simple harmonic oscillation the displacement is given by x=15sin(Ït3) where t is in seconds. What are the angular frequency and period for this motion?
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WORKED EXAMPLE 12.3
The graph of Figure 12.20 shows the variation with displacement x of the acceleration a of a particle.
aExplain why the oscillations are simple harmonic.
bDetermine the period of oscillation.
cDetermine the maximum speed and acceleration in this motion.

Figure 12.20: For worked example 12.3.

Answer
aThe oscillations are simple harmonic because the acceleration is proportional to displacement (straight-line graph through the origin) and opposite to it (negative gradient). So it fits the defining equation a = âÏ2x.
bThe gradient of the line in the graph is âÏ2. From the graph the gradient is:
âÏ2Ï2Ï=â3.0Â mÂ sâ20.12Â m=25Â sâ2=5.0Â rad sâ1.
Hence the period is:
T=2ÏÏ=2Ï5.0â1.3Â s.
cThe maximum speed is Ïx0, i.e. 5.0 Ã 0.06 = 0.30 m sâ1. The maximum acceleration is Ï2x0, i.e. 25 Ã 0.06 = 1.5 m sâ2.
EXAM TIP
Watch the unit conversion. The amplitude is read from the graph as 0.06 m.
WORKED EXAMPLE 12.4
A particle undergoes SHM with amplitude 4.0 mm and frequency 0.32 Hz. At t = 0, the displacement is 4.0 mm.
aWrite down the equation giving the displacement and velocity for this motion.
bCalculate the velocity when the displacement is 2.0 mm for the first time.
Answer
aWe start with x = x0 sin(Ït + Ï).
We are told that at time zero 4.0 = 4.0 sin(0 + Ï)
So the phase angle is Ï2.
Hence x=4.0Â sin(Ït+Ï2)=4.0Â cos(Ït).
We have been given frequency f so the angular frequency Ï is:
Ï=2Ïf=2ÏÃ0.32=2.01â2.0Â rad sâ1.
So the equation for the displacement is:
x=4.0cos(2.0t)
where t is in seconds and x is in millimetres.
The equation for the velocity is:
v=Ïx0cos(Ït+Ï)=8.0cos(2.0t+Ï2)=â8.0sin(2.0t).
bFrom x = 4.0 cos (2.0t) we find 2.0 = 4.0 cos (2.0t) and so cos(2.0t)=12. This means that 2.0t=Ï3ât=0.5236Â s. At this time the velocity is v = â8.0 sin(2.0 Ã 0.5236) = â6.9 mm sâ1.
(Make sure your calculator is set in radians.)
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WORKED EXAMPLE 12.5
The graph of Figure 12.21 shows the displacement of a particle from a fixed equilibrium position.

Figure 12.21: For worked example 12.5.

aUse the graph to determine:
ithe period of the motion
iithe maximum velocity of the particle during an oscillation
iiithe maximum acceleration experienced by the particle.
bOn a copy of the diagram, mark:
ia point where the velocity is zero (label this with the letter Z)
iia point where the velocity is positive and has the largest magnitude (label this with the letter V)
iiia point where the acceleration is positive and has the largest magnitude (label this with the letter A).
Answer
aiThe period is read off the graph as T = 0.20 s.
Since T=2ÏÏ we have that:
Ï=2ÏT=31.4â31Â radÂ sâ1
iiThe maximum velocity is then:
vmax=Ïx0=31.4Ã0.020=0.63Â mÂ sâ1
iiiThe maximum acceleration is found from:
amax=Ï2x0=31.42Ã0.020=20Â mÂ sâ2
biThe velocity is zero at any point where the displacement is at a maximum or a minimum.
iiFor example at t = 0.15 s.
iiiFor example at t = 0.10 s or t = 0.30 s.
CHECK YOURSELF 8
In a simple harmonic oscillation the maximum speed is 4.0 m sâ1 and the maximum acceleration is 8.0 m sâ2. What is the period?
TEST YOUR UNDERSTANDING
15The restoring force depends on displacement from equilibrium x through
âF=k2x
âF=âk2sinx
âF=k2xxâ1
âF=(1âek2x).
aExplain why none of these gives rise to simple harmonic oscillation.
bFor students who know about MacLaurin series: are any of these simple harmonic for small amplitude?
16A 2.0 kg block is connected to two springs. Each spring has spring constant k = 120 N mâ1. The springs have their natural length.

Figure 12.22: For question 16.

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aThe springs are connected as in part a of Figure 12.22. Calculate the period of the oscillations of this block when it is displaced from its equilibrium position and then released.
bThe springs are now connected as in part b of the diagram. State and explain whether the period changes.
17In a SHM the maximum velocity is 12 m sâ1 and the maximum acceleration is 48 m sâ2. What is:
athe period of oscillations
bthe amplitude of oscillations?
18In simple harmonic motion x = x0 sin(Ït + Ï) and v = Ïx0 cos (Ït + Ï).
aUsing the identity sin2 Î¸ + cos2 Î¸ = 1 for Î¸ = Ït + Ï, show that v2=Ï2(x20âx2).
bThe maximum speed in a simple harmonic oscillation is v. What is the speed when the displacement is half the amplitude?
19Figure 12.23a shows a rectangular solid of mass M floating in a liquid of density Ï. The cross sectional area of the solid is A. In Figure 12.23b the block has been pushed downward by a distance x.

Figure 12.23: For question 19.

aShow that SHM will take place when the block is released.
bShow that the period of oscillations is given by T=2ÏMÏgAâââââ.
20Figure 12.24 shows a simple pendulum of length L. The mass of the bob is m.

Figure 12.24: For question 20.

aDraw the forces on the bob when the pendulum has been displaced by an angle Î¸ to the vertical.
bShow that the acceleration of the bob in this position is a = âg sin Î¸.
cThe displacement x of the bob is the length of the arc subtended by the angle Î¸ and x = LÎ¸.
Explain why for very small angles in radians the acceleration is approximately a=âgLx.
dHence deduce that the period of oscillations is given by T=2ÏLgâââ.
21A point undergoes SHM oscillations with amplitude 5.0 cm and period 0.25 s. Calculate, for this point,
athe maximum speed
bthe maximum acceleration.
22The displacement, in metres, in a simple harmonic motion is given by x=6.0sin(2t3+Ï5) where t is in seconds. What are the angular frequency and period?
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23The displacement of a particle executing SHM is given by x = 5 sin(2Ït) where x is in millimetres and t is in seconds. Calculate:
athe initial displacement of the particle
bthe period of the oscillations
cthe displacement at t = 1.2 s
dthe time at which the displacement first becomes â2.0 mm.
24A particle undergoes SHM oscillations such that the displacement, in cm, is given by x=12sin(Ït2), where time is in seconds. Calculate, for this particle, when the displacement is 6.0 cm:
athe velocity
bthe acceleration.
25aWrite down an equation for the displacement of a particle undergoing SHM with amplitude 8.0 cm and frequency 14 Hz, assuming that at t = 0 the displacement is 8.0 cm and the particle is at rest.
bFind the displacement, velocity and acceleration of this particle at a time of 0.025 s.
26A point on a guitar string oscillates in SHM with an amplitude of 5.0 mm and a frequency of 460 Hz. Determine the maximum velocity and acceleration of this point.
27A guitar string, whose two ends are fixed, oscillates as shown in Figure 12.25.

Figure 12.25: For question 27.

The vertical displacement of a point on the string a distance x from the left end is given by y = 6.0 cos(1040Ït) sin(Ïx), where y is in millimetres, x is in metres and t is in seconds. Use this expression to:
adeduce that all points on the string execute SHM with a common frequency and common phase, and determine the common frequency
bdeduce that different points on the string have different amplitudes
cdetermine the maximum amplitude of oscillation
dcalculate the length L of the string
ecalculate the amplitude of oscillation of the point on the string where x=L4.
28Figure 12.26 shows the variation with time t of the velocity v of a particle executing SHM.

Figure 12.26: For question 28.

aEstimate the area under the curve from 0.10 s to 0.30 s.
bWhat does this area represent?
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29Figure 12.27 shows the variation with time t of the displacement x of a particle executing SHM.

Figure 12.27: For question 29.

Draw a graph to show the variation with displacement x of the acceleration a of the particle (put numbers on the axes).
30A body of mass 0.120 kg is placed on a horizontal plate. The plate oscillates vertically in SHM making five oscillations per second (see Figure 12.28).

Figure 12.28: For question 30.

aDetermine the largest possible amplitude of oscillations such that the body never loses contact with the plate.
bCalculate the normal reaction force on the body at the lowest point of the oscillations when the amplitude has the value found in a.
31What is the phase angle for each of the two simple harmonic oscillations shown in Figure 12.29?

Figure 12.29: For question 31.

32Show that the phase angle is given by the formula Ï=arctanÏx(0)v(0) where x(0) and v(0) are the displacement and velocity at time zero.
33Consider a particle that rotates counterclockwise around a circle of radius R with constant angular speed Ï. At time zero, the particle is in the position shown, making an angle Ï with the positive x-axis.

Figure 12.30: For question 33.

Show that the x and y coordinates of the particle perform simple harmonic oscillations.
