S6.3 The SHM equation: a general tour
Preamble
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Preamble
- We consider the general properties of the SHM equation (Key Point 6.2)
- We prepare for the mathematics by considering the SHM equation qualitatively.
- The key features of this equation of motion are that
depends on x –we have previously considered only situations where the force driving the motion is constant, at least in magnitude.
is always opposite to x, back towards the equilibrium position
is linearly proportional to x, vanishing at x=0
the motion must conserve energy (section S3.7).
- Energy-conservation tells us that
- The coordinate x cannot grow arbitrarily big – this would entail gaining some energy
- x cannot simply come to rest at x=0 –this would entail losing energy
- We are left only with oscillatory motion about x=0 – which is what we would expect from our experience of near stable equilibrium systems
[A] The solution: algebra
- The mathematical problem is to solve the SHM equation (Key Point 6.2).
- This is a differential equation; solving it means finding the general form of a function x(t) which satisfies it.
- Not all differential equations are solvable in this way; but this one is.
- It is solvable (easily) because it is linear –it contains no power of x beyond the first:
Key Point 6.3
The general solution to the SHM equation
x(t)=xmcos(ωt+φ)
where xm and φ are constants determined by the initial conditions.
Commentary
- To show that
x=xmcos(ωt+φ)is a solution of the SHM equation we differentiate it with respect to time giving the speed:
- We differentiate again to give the acceleration
where the last step uses the proposed solution again and recovers the SHM equation.
- Since the proposed solution works (satisfies the SHM equation) for any choices of xm and φ it is the general solution to this equation.
- These two constants are fixed by the initial conditions …whatever is happening at the time we call t=0: we will come back to this shortly
In this system, knowledge of the initial conditions allows us to predict the future
Should we believe that promise?
The short answer is that we should (at least) be cautious; for a longer answer see S6.7
[B] Visualisation: the role of amplitude and phase
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- Suppose that at t=0 this vector makes an angle φ with the x-axis.
- Then at time t this angle is ωt+φ
- The x coordinate of the particle is the projection of
the radial vector on the x axis:
x=xmcos(ωt+φ)which is the SHM solution.
[C] Visualisation: displacement, velocity and acceleration
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- At the equilibrium position, x=0, the acceleration is zero but the speed is maximal.
- At the extremes of motion x=±xm the speed is zero but the acceleration is maximal.
The velocity is π/2 out of phase with the displacement.
[D] How initial conditions fix amplitude and phase
- The initial conditions for any equation of motion are defined by two quantities, the displacement and the velocity.
- These two quantities x(0) and
together define the amplitude xm and the initial phase φ.
- The relationships are expressed in the equations:
Check It!
- Are the units OK?
- Does it make sense?
[E] All about the ’frequency’
- The physical quantity represented by ω in the above equations is called the angular frequency of the motion.
- The analogy between SHM and circular motion (see above) explains this terminology.
- There are alternative ways of expressing the same thing:
The frequency is defined by
and gives the number of cycles completed each second
The period is defined by
and gives the time required to complete one cycle.

Commentary
- It is helpful to see how the relationship between T and ω arises.
- Recall the equation for the displacement
x(t)=xmcos(ωt+φ)
- Then deduce that a time T later
the displacement is
x(t+T)=xmcos(ω(t+T)+φ)=xmcos(ωt+ωT+φ)
- Setting T=2π/ω then gives
x(t+T)=xmcos(ωt+2π+φ)=xmcos(ωt+φ)=x(t)
- Thus waiting for T=2π/ω takes us back to the same displacement
- The same argument applies to
and
(velocity and acceleration) which are also sine or cosine functions of ωt+φ
- Waiting for T thus restores the status quo completely.
- Note the units of the three quantities:
quantity symbol units angular frequency ω rad⋅s-1 frequency f s-1≡Hz period T s - Finally note the fundamental differences
- xm and φ are fixed by initial conditions –ie how we start the system off
(displaced but from rest? with a push from the equilibrium position? ) - ω and T are fixed by the physical system itself
– what kind of system it is (spring, pendulum, drum …)
– its particular physical parameters (spring constant, length of support, tautness …)
- xm and φ are fixed by initial conditions –ie how we start the system off
- In particular T is
independent of the amplitude of the motion, xm
Think about it!
This is a surprsing result. It reflects a neat little conspiracy. If you change the amplitide of a simple harmonic oscillator you do two things: you change the distance it has to cover in its cycle; but you also change the energy you are giving it to accomplish the trip. The one change exactly offsets the other!
Learning Resources
![]() | HRW 15.3 |
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