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Chapter 13: Simple Harmonic Motion and Hooke’s Law Please remember to photocopy 4 pages onto one sheet by going A3→A4 and using back to back on the photocopier.
Hooke’s Law Hooke’s Law states that when an object is stretched the restoring force is directly proportional to the displacement, provided the elastic limit is not exceeded.
We represent this mathematically as follows; F ∝ - s* ⇒
The most common example of Hooke’s Law is a stretched string.
Simple Harmonic Motion (S.H.M.) An object is said to be moving with Simple Harmonic Motion if;*
- its acceleration is directly proportional to its distance from a fixed point in its path, and
- its acceleration is directed towards that point.
We represent this mathematically as :
- The ‘fixed point’ referred to in the definition is also called the ‘equilibrium position’.
- The key to SHM is that the acceleration and the displacement are always opposite in direction.
- This is represented by the minus sign in the equation, although mathematically the minus sign has no significance and can be ignored for calculations. “ But I thought ω represented angular velocity?” *
Other examples of SHM, apart from a stretched string are:
- A prong on a vibrating tuning fork,
- A pendulum oscillating at a small angle,
- The tides.
To show that any object that obeys Hooke’s Law will also execute SHM The aim here is to show that Hooke’s Law is a subset of S.H.M., i.e. that F ∝ -s can be rewritten in the form a = - ω^2 s.
So we start with the equation for Hooke’s Law; F ∝ -s ⇒ F = -k s But F = ma ⇒ ma = -k s
Now divide both sides by m ⇒ a = - s This is equivalent to the equation for S.H.M. where the constant ω^2 in this case is k/m.
Relationship between Periodic Time (T) and ω for an object undergoing SHM* As with waves, the periodic time is the time taken for one complete oscillation. Also T = 1/f, where f is the frequency. Remember ω is not angular velocity. What is the unit of ω in this particular case?
The Simple Pendulum A simple pendulum consists of a light string with a mass at the end. For an angle of swing less than approximately 5^0 , the motion of a Simple Pendulum can be considered to be SHM.
Periodic Time of a Simple Pendulum
a = - ω^2 s
F = - k s
g T = 2 π l
Mandatory Experiment: Investigation of the relationship between period and length for a simple pendulum, and hence calculation of g.
Leaving Cert Physics Syllabus
Content Depth of Treatment Activities STS
S.H.M. and Hooke’s law.
Hooke’s law: restoring force ∝ displacement. F = -ks ma = -ks a = = -ω^2 s
Demonstration of SHM, e.g. swinging pendulum or oscillating magnet.
Appropriate calculations.
Everyday examples.
Systems that obey Hooke’s law e.g. simple pendulum, execute s.h.m. T =
*Extra Credit F ∝ -s Why the minus sign? Well because displacement is a vector, it has a direction associated with it, and this direction is always opposite to the direction of the restoring force, even though as one quantity gets bigger so does the other.
*Simple Harmonic Motion (S.H.M.) Basically what’s going on is that an object has its natural resting point, and if, when it gets disturbed from this point it tries to return but actually overshoots the mark and has to return again and again, then the object is executing SHM.
In our study of Newton’s Laws of Motion we consider the motion of a body when subject to a constant force or forces. As a result we can calculate the object’s velocity or position at any time.
However there are many instances when a moving object is subject to a changing force – can we still calculate future position and velocity? Well, we can if we can quantify the force, i.e. if we know how the force is changing. One example of an object experiencing a changing force is a stretched spring. We say that the spring is undergoing Simple Harmonic Motion (SHM).
What happens acceleration in Simple Harmonic Motion? Recap
- In linear motion, acceleration is always constant in direction and magnitude
- In SHM, acceleration by its nature is not constant in either direction or magnitude.
Let’s take another look at this. Consider a Simple Pendulum* displacement velocity acceleration Equiplibrium position
0 max 0
Extreme position
max 0 max
The aspect which most people find confusing here is that at the extremities the velocity is zero, while the acceleration is a maximum, whereas at the equilibrium position the opposite is the case. How can this be? Answer: You must think in terms of forces Personally, I can’t fathom why textbooks don’t explain this by referring to forces; after all, the only way you can have acceleration is if you have a force causing it. Acceleration is directly proportional to the force causing it, so if force is a maximum acceleration must be also, and vice versa. Now let’s look at the diagram in terms of forces. At the equilibrium point the restoring force on the mass is zero – it’s not being pulled to the left or the right, therefore the acceleration must be zero, whereas at the extremities the force pulling the mass back in is a maximum, therefore the corresponding acceleration must also be also be a maximum.
The other point which might cause confusion is velocity being zero at the extreme positions, but remember we came across this before; when an object is thrown up its instantaneous velocity is zero at the very top, (even though its acceleration is not zero).
*** “** But I thought ω represented angular velocity?” Well spotted. ω is the same symbol as was used in the last chapter (Circular Motion) to represent angular velocity, however it does not represent angular velocity here. It’s not just coincidence that ω turns up in both chapters; they are connected, but the relationship is a little complicated.
In the last chapter we made use of the formula v = distance/time and applied it to a satellite undergoing a full cycle where v = 2πr /T. We also had v = rω, so we can re-write the previous equation as rω = 2π r/T. Cancel r’s to get ω = 2π /T. Cross-multiply to get T = 2π/ω. Hmmm... - what does a stretched string have to do with circular motion
