Question 6: Simple Harmonic Motion

Question 6: Simple Harmonic Motion

Please remember to photocopy 4 pages onto one sheet by going A3→A4 and using back to back on the photocopier

Page / Commencement date / Questions covered
Introduction
Formulae
Manipulation of equations
Questions involving ε
Introduction to Problem-Solving questions
Vertical Motion Exam Questions
Horizontal Motion Exam Questions
Guide to answering Exam Questions

************* Marking Schemes / Solutions to be photocopied separately ***********

With the exception of the ‘Manipulation of equations’ section, all other questions will involve the following series of steps and so they need to be learned off.

I am just placing this here to make it easy to find the information when you need it – you don’t need to learn this now.

The extensionof a spring or a string is the distance between the object’s Current Positionand the end of the natural length of the string (in other words, how much the string is extended beyond its natural length).

The extension depends on the position of the mass, which changes throughout the question.

Or
The extension continually changes as the mass moves over and back.

The Amplitude is the distance from the Release Point (R.P.) to the Equilibrium Position (E.P.).
The amplitude therefore does not change.

N clucking B:
The release point has no bearing on the question until after you have established step 3 (i.e. shown SHM). This is probably the most common mistake that students make when answering these questions.

Procedure for solving long questions

Find Equilibrium Position (E.P.)
If we’re working in the horizontal direction with a single string then equilibrium position is at the end of the string.

If we’re working in the vertical direction then we use the fact that at E.P.,Fdown = Fup

I tend to use the letter ‘d’ to represent this extension so that it doesn’t get confused with ‘x’ which we will use in the next section.

Show SHM

If we’re working in the horizontal direction (assuming just one string and no friction) then there is no small force.
The big force will be the tension in the string so we can use F{ext) to represent it.

So for a string in the horizontal direction our equation will be0 – k {ext} = ma.

If we’re working in the vertical direction then the small force will be mg

So for a spring in the vertical direction our equation will bemg – k {ext} = ma.

Before answering long questions check that you understand this by testing yourself on the last page.

Introduction

Basically what’s going on is that an object which exhibits simple harmonic motion has its natural resting point (called ‘equilibrium position’) and when it gets disturbed from this point it heads back but overshoots the mark and has to return again and again.

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 timet.

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, 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).

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. We will ignore it from now on.

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

What happensacceleration in Simple Harmonic Motion?

Recap

With vuast, acceleration is constant inmagnitude (for a given stage).
In SHM, acceleration by its nature is not constant in 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 emphasise this more. 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 above 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?”

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 chapter on Circular Motion 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.

Formulae

a = acceleration

A = Amplitude

x = distance from equilibrium position

 = constant

v = velocity at any time t

T = periodic time (time for one complete oscillation)

Know the following equations

a = 2x / Basic equation which defines simple harmonic motion
v2 = 2(A2 – x2)
x = A Sin t
/ (if the clock starts (t = o) as the particle is passing the equilibrium position)
x = A Cos t / (if the clock starts when the particle is at the extreme position – but x still represents displacement from the equilibrium position, not the extreme point)
amax = 2A / Can you see how we get this from the first equation above?
vmax = A / Can you see how we get this from the second equation above?
T= /
Where f = frequency = the number of oscillations per second.

Manipulation of equations

Notes

The time to go through one complete oscillation is the period T.
The time to go from one extreme to the other corresponds to the time to go through half an oscillation, and this equals T/2.
The time to go from the release point to the equilibrium position corresponds to the time to go through one amplitude, and this equals T/4.

However at any other sub-intervals the distance travelled is not directly proportional to the time, and this is why we revert to the equation: x = A Sint etc.

Also with these equations, x always corresponds to the displacement between the given position and the equilibrium position.

This can cause confusion if the object is released from an extreme and the formula: x =A Cost is used.

Know the significance of the phrase ‘when the string is taut’, and ‘when the string is slack’ (when the string is taut the object will be undergoing SHM, but not when it is slack).

Quite often the object will only be exhibiting SHM for part of its motion; you must be able to figure out when it is and when it is not exhibiting SHM.

It will only be exhibiting SHM when the string is stretched.

You should be able to use VUAST for the other situations.

The approach is VERY similar to how you answer vuast-type questions:

Write down what you know in terms of the variables
Write down all the equations to remind you of your options
Identify the equation you need
Solve

OMG, I can’t remember the difference between extension and amplitude!!!

This is a killer. Picture the situation; a string has got a natural length of 1 m.

The string is fixed at one end and there is a mass of 3 kg at the other end.

The string is now pulled a further 2 metres and is then released.

The extension is the distance between the object’s Current Position (C.P.) and the end of the natural length of the string (in other words, how much the string is extended by from its natural length).

The extension continually changes as the mass moves back.

The Amplitude is the distance from the Release Point (R.P.) to the Equilibrium Position (E.P.).

In these situations the equilibrium position corresponds to the point at the end of the string’s natural length.
Why? Two reasons (at least); it is the point beyond which the mass will exhibit SHM if released, and secondly it also corresponds to the point of maximum velocity (and zero acceleration) as the mass passes back through.

In the scenario above the extension changes with time but the amplitude is fixed at 2 m throughout the question, regardless of where the mass is.

Let’s go

2003 (a)

A particle is moving with simple harmonic motion of period π seconds about a fixed point o.

The maximum speed of the particle is 8 cm/s.

(i)Find the amplitude of the motion.

(ii)Find the speed of the particle when it is at a distance of 3 cm from o.

1979 (a)

A body is moving with simple harmonic motion, of amplitude 5m.

When it is 4m from the midpoint of its path its speed is 6 m/s.

Find its speed when it is 2.5m from the mid-point.

2012 (a)

A particle of mass 0·5 kg is suspended from a fixed point P by a spring whichexecutes simple harmonic motion with amplitude 0·2 m.

The period of the motion is 2 seconds.

Find

(i)the maximum acceleration of the particle

(ii)the greatest force exerted by the spring correct to one place ofdecimals.

2001 (a)

A particle moving with simple harmonic motion has speeds of 5 cm/s and 2 cm/s when it is at points 3 cm and 4 cm, respectively, from the centre of the motion.

(i)Find the amplitude and the period of the motion.

(ii)Find the maximum speed of the particle.

1996 (a)

A body of mass 10kg moves with simple harmonic motion.

At a displacement of 0.8 m from the centre of oscillation, the velocity and acceleration of the body are 2 m/s and 20 m/s2 respectively.

Find

(i)the number of oscillations per second
{for some reason the marking scheme assumed you were asked for the number of oscillations per minute – don’t why they made this mistake}

(ii)the amplitude of motion

(iii)the maximum acceleration and hence show that the force to overcome the inertia of the body at the extremity of the oscillation is 223.6 N.

1993 (a)

A particle of mass m moves with simple harmonic motion under the action of a variable force.

If the maximum value of the force is and the amplitude of the motion is 4 m calculate

(i)the period of the oscillation

(ii)the speed of the particle at a time seconds after passing through the centre of oscillation.

2010 (b)

A particle moves with simple harmonic motion of amplitude 0.75 m.

The period of the motion is 4 s. Find

(i)the maximum speed of the particle

(ii)the time taken by the particle to move from the position of maximum speed to a position at which its speed is half its maximum value.

1990 (a)

(i)A particle starts from rest, and moves with simple harmonic motion of period 6n seconds.

(ii)Show that the particle moves from the position of maximum velocity to the position in which the velocity is half the maximum in n seconds.

2006 (a)

A particle moves with simple harmonic motion of period 3π.

At time t = 0, the particle passes through the centre of the oscillation.

It passes through a point distant 4 m from the centre of motion with a speed of 5 m/s away from the centre.

Find, correct to two decimal places,

(i)the maximum acceleration of the particle

(ii)the time which elapses before it next passes through this point.

1986(a)

A particle moves with simple harmonic motion through two pointsp and q 1.2m apart.

Its velocity is the same at p as at q. It takes 3seconds to move from p to q and 3 seconds to move from q to q i.e. passing q the next time.

Using a diagram or otherwise find the period of the particle and the amplitude.

1980 (b)

A particle is moving in a straight line with simple harmonic motion.

When it is at a point p1 of distance 0.8m from the mean-centre, its speed is 6m/s and when it is at a point p2 if distance 0.2m from the end position on the same side of the mean-centre as p1, its magnitude is of magnitude 24m/s2.

Ifr is the amplitude of the motion, show that and hence find the value of r.

Find also the period of the motion and the shortest time taken between p1 and p2 correct to two places of decimals.

1981 (a) {ODDBALL}

Establish the formula T = for the periodic time of a simple pendulum of length l.

The length of a seconds pendulum (T = 2secs) is altered so as to execute 32 complete oscillations per minute.

Calculate the percentage change in length.

Show that the particle is exhibiting simple harmonic motion

You need to have calculus (differentiation) covered in maths before you tackle this particular section.

Start with x = a sin ωt and differentiate twice.

1983 (a)

Define simple harmonic motion in a straight line and show that x = asinωt can describe such motion, when x is the distance from a fixed point and a, ω and t have the usual meanings.

1980 (a)

A particle is moving in a straight line such that its distance x from a fixed point at time t is given by

x = r cos ωt.

Show that the particle is moving with simple harmonic motion.

1998 (a)

Define Simple Harmonic Motion. The distance, x, of a particle from a fixed point, o, is given by:

x = 7 sin ωt + 24 cos ωt, ω being a constant.

(i)Show that the particle is describing simple harmonic motion about o.

(ii)Calculate the amplitude of the motion.

1992 (a)

If the displacement of a moving particle at any one time t is given by the equation x = 5Cosω t + 12Sinωt

(i)show that the motion is simple harmonic motion.

(ii)calculate the amplitude of the motion.

Questions involving ε

If the clock starts when the particle is at an angle  from Equilibrium Position (i.e. in between the equilibrium and the extreme point) then the motion of the particle is described by the equation:

x = A Sin (t + ) where  represents the angle at time t = 0

Remember that angles are measured in radians

1974 (a)

Define Simple Harmonic Motion in a straight line and show that if the magnitude of the displacement from the equilibrium position after time t is given by x = asin(ωt + α), where a, ω, α are constants, then the notion is simple harmonic.

2011 (a)

The distance, x, of a particle from a fixed point, O, is given by

ε)

wherea, ε andωare positive constants.

(i)Show that the motion of the particle is simple harmonic.

(ii)A particle moving with simple harmonic motion starts from a point 1 m from the centreof the motion with a speed of 9.6 m s-1 and an acceleration of 16 m s-2.

Calculate a,ω and ε.

2009 (a)

The distance, x, of a particle from a fixed point, o, is given by x = a cos (ωt+ε ) where a, ω and ε are constants.

(i)Show that the motion of the particle is simple harmonic.

(ii)A particle moving with simple harmonic motion starts from a point 5 cm from the centre of the motion with a speed of 1 cm/s. The period of the motion is 11 seconds.

Find the maximum speed of the particle, correct to two decimal places.

1982 (b)

The distance, x, of a particle from a fixed point, o, is given by x = acos(ωt + α) where a, ω, α are positive constants.

(i)Show that the particle is describing simple harmonic motion about o and calculate ω and α if the velocity v = –2a and x = when t = 0.

(ii)After how many seconds from the start of the motion is x = 0 for the first time?

(See Tables P.8. Take pi = 3.142).

1999 (a)

A particle moves with simple harmonic motion of period .

Initially it is 8 cm from the centre of motion and moving away from the centre with a speed of 42 cm/s.

Find an equation for the position of the particle in time t seconds.

2004 (b)

A particle moves in a straight line such that its displacement from a fixed point o at time t is given by

x = a cos(ω t − β) where a, ω and β are positive constants.

(i)Show that the motion of the particle is simple harmonic motion.

(ii)The period of the motion is 16 seconds. At time t = 4 s, the particle is 12 m from o and 4 s later the particle is on the other side of o and at a distance of 5 m from o.

Find a, ω and β.

Introduction to Problem-Solving type questions in the x-direction

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 - extension

The minus sign indicates that the restoring force and the extension are in opposite directions.



Or F = - k x

The most common example of Hooke’s Law is a stretched string.

When using the F = -k x expression for Hooke’s Law, x represents the extension, i.e. the distance between the new length and the original (natural) length.

Wait; I’ve forgotten the difference between extension and amplitude!!!

This is a killer. Picture the situation; a string has got a natural length of 1 m.

The string is fixed at one end and there is a mass of 3 kg at the other end.

The string is now pulled a further 2 metres and is then released.

The extension continually changes as the mass moves back up.

The Amplitude is the distance from the Release Point (R.P.) to the Equilibrium Position (E.P.).

In the scenario above the extension changes with time but the amplitude is fixed at 2 m throughout the question, regardless of where the mass is.