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These are the Lecture Notes of Applied Maths which includes Initial Velocity, Ordinary Level, Horizontal Surface, Right Angles, Inclined Plane, etc. Key important points are: Simple Harmonic Motion, Manipulation, Involving, Problem Solving, Vertical Motion, Horizontal Motion, Natural Resting Point, Return Again and Again, Changing Force, Simple Harmonic Motion
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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
Introduction 2
Formulae (^) 3
Manipulation of equations 4
Questions involving ε 7
Introduction to Problem-Solving questions 8
Vertical Motion Exam Questions 10
Horizontal Motion Exam Questions 14
Guide to answering Exam Questions 16
*************** Marking Schemes / Solutions to be photocopied separately *************
Introduction
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).
An object is said to be moving with Simple Harmonic Motion if;
We represent this mathematically as :
Other examples of SHM, apart from a stretched string are:
What happens acceleration in Simple Harmonic Motion? Recap
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? (see next page for the answer)
a = - ω^2 x
Manipulation of equations 1983 (a) Define simple harmonic motion in a straight line and show that x = a sin ω t 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 ω t. 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 = 5 Cos ω t + 12 Sin ωt (i) show that the motion is simple harmonic motion. (ii) calculate the amplitude of the motion.
2012 (a) A particle of mass 0·5 kg is suspended from a fixed point P by a spring which executes 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 of decimals.
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.
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.
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.
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 10 kg 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/s^2 respectively. Find (i) the number of oscillations per second (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 16
7 m 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 7
seconds after passing through the centre of oscillation.
1990 (a) A particle starts from rest, and moves with simple harmonic motion of period 6 n seconds. 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.
1986 (a) A particle moves with simple harmonic motion through two points p and q 1. 2 m apart. Its velocity is the same at p as at q. It takes 3 seconds 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 p 1 of distance 0.8 m from the mean-centre, its speed is 6 m/s and when it is at a point p 2 if distance 0.2 m from the end position on the same side of the mean-centre as p 1 , its magnitude is of magnitude 24 m/s^2.
If r is the amplitude of the motion, show that
64
2 3
2 (^2) −
r
r and hence find the value of r.
Find also the period of the motion and the shortest time taken between p 1 and p 2 correct to two places of decimals.
Questions involving ε
equilibrium and the extreme point) then the motion of the particle is described by the equation
x = A Sin ( ω t + ε ) where ε represents initial displacement 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 = a sin(ω t + 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) where a , ε and ω are 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 centre of 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 = a cos (ω t + 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 = –2 a and x = 5
3 a (^) 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
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. A mass hangs from it and as a result the string now is 1.5 m in length. The string is next pulled down 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 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.
To begin with, it doesn’t matter where the object is released from!!! This is because the amplitude only comes into play towards the end of the question. You will usually be told where the particle is released from in the question, but ignore this information completely until after you have first demonstrated that the particle is exhibiting S.H.M.
Steps
1. Find the position of equilibrium Forces left = Forces right, or Forces up = Forces Down. This gives a value for d, and hence a position of equilibrium. If however, you are dealing with a string which lies horizontally on a table, then simply take the equilibrium position point to be at the end of the natural length of the string. **If we refer to this extension as ‘d’ it avoids confusion with ‘x’ which we should be using in the next step.
Vertical Motion exam questions 2008 (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically. The elastic constant of the string is 500 N/m. The mass is pulled down a vertical distance of 20 cm from the equilibrium position and is then released from rest. (i) Show that the particle moves with simple harmonic motion. (ii) Find the speed and acceleration of the mass 0.1 seconds after it is released from rest.
1998 (b) An elastic string of natural length one metre is extended 20cm by a particle attached to its end and hanging freely. The particle is then pulled down a further distance of 40cm and released. Show that the particle moves with simple harmonic motion when the string is taut. Find the height above the equilibrium position to which the particle will rise.
2007 (a) A particle of mass m kg is suspended from a fixed point p by a light elastic string. The extension of the string is d when the particle is in equilibrium. The particle is then displaced vertically from the equilibrium position a distance not greater than d and is then released from rest. (i) Show that the motion of the particle is simple harmonic. (ii) Find, in terms of d , the period of the motion.
2001 (b) A particle of mass m kg is suspended from a fixed point p by a light elastic string of natural length l and
elastic constant 4 mg l
(i) Find the distance of the equilibrium position from the point p, in terms of l.
(ii) The particle is pulled down until it is at a distance
l vertically below p and is then released from rest.
Find the time taken, in terms of l, for the string to go slack.
1991 (b) A particle of mass 3 kg is suspended by a light string which is found to extend by 20 cm when the particle is at rest. If the upper end of the string is held firm and the particle is pulled down slightly and then released, find the period of the resultant motion.
Define simple harmonic motion. A body of mass 0. 25 kg hangs from a spiral spring. When pulled down 10 cm below its equilibrium position and released, it vibrates with simple harmonic motion of period 2 seconds. (i) Find its velocity as it passes through the equilibrium position. (ii) What is the shortest time taken to travel from a point 2 cm below the position to a point 2 cm above the equilibrium position? (iii)Find the elastic constant of the spring. (iv) By how much will the spring shorten when the body is removed?
Define simple harmonic motion. A particle of mass m is suspended from a fixed point p by a light extensible string of natural length d and
elastic constant d
49 m .
It is pulled down a distance 5
8 d below p and is then released from rest.
(i) Show that the particle moves with simple harmonic motion as long as the string remains taut. (ii) Find in terms of d , when the string becomes slack for the first time.
2000 (b) A particle of mass 0.3 kg is attached to the midpoint of a light elastic string of natural length 1 m and elastic constant k. The string is then stretched between two points a and b. The point a is 2 m vertically above b. (i) Find the extensions of the two parts of the string, in terms of k, when the system is in equilibrium (ii) Find the minimum value of k which will ensure that the lower part of the string is taut (iii)Find the period of small oscillations, in terms of k, when the particle is displaced vertically. (Assume both parts of the string remain taut.)
Define simple harmonic motion. A particle of mass 2 kg is attached to the ends of two elastic strings, each of natural length 1 m and elastic constant 49 N/m. The other ends of the two strings are attached to two fixed points a and b in the same vertical line where a is 4 m above b. The particle is released from rest from the midpoint of ab. By considering the forces acting on the particle when it is x metres from a , where 2 < x < 2.4, show that it is moving with simple harmonic motion. Find the least time taken for the particle to reach the point x = 2.3, and find its speed there.
1997 no.7 (a) A particle of weight W is attached to two inextensible strings each of length 26 cm. The other ends of the strings are attached to two points on the same horizontal level 48 cm apart. Find the tension in each of the strings in terms of W. The strings are replaced by two light elastic strings, each of natural length 26 cm and elastic constant k , attached in the same way. The position of equilibrium of the particle is now 8 cm below its previous position of equilibrium. Find the value of k in terms of W.
1986 (b)
An elastic string of natural length 2 a is stretched between two points which are 2 3 a apart. A particle of weight W is attached to its midpoint and hangs in equilibrium with the string inclined at an angle of 30^0 to the horizontal. Show that the modulus of elasticity of the string is W.
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 = a sin(ω t + α), where a , ω, α are constants, then the notion is simple harmonic. A light flexible elastic string pq , of natural length 1 m and elastic constant 245 N/m, has one end p tied to a fixed point and has a particle of mass 5 kg attached to the end q. The particle is held 1 m below p and is then released from rest to fall under gravity. By considering the forces acting on the particle when it has fallen a distance (0·2 + x ) m, show that it moves with simple harmonic motion and that its acceleration is zero when x = 0. Find the time taken to fall a distance 0·3 m.
Horizontal Motion exam questions 2005 (b) A light elastic string of natural length a and elastic constant k is fixed at one end to a point o on a smooth horizontal table. A particle of mass m is attached to the other end of the string. Initially the particle is held at rest on the table at a distance 2a from o, and is then released.
Show that the time taken for the particle to reach o is.
1996 (b) A light perfectly elastic string of natural length a and elastic constant k is fastened at one end p to a fixed point of a smooth horizontal table, and a particle of mass m is attached to the other end. The particle is held on the table at a distance 2 a from p and then released. Prove (i) that the particle executes simple harmonic motion while the string is taut.
(ii) that the particle reaches p after k
m
seconds.
1992 (b) A particle of mass 2 kg is attached to one end of a light elastic string of natural length 1 m and elastic constant 14 N/m. The other end of the string is fixed to a point A on a smooth horizontal table. The particle is pulled across the table and released from rest at a point C which is a distance 1. 5 m from A. If B is a point on AC such that |AB| = 1 m, (i) prove that the particle performs simple harmonic motion when travelling from C to B. (ii) calculate the time taken to travel from C to B.
(iii)prove that the particle then travels for 7
s with constant speed.
1999 (b) A smooth particle of mass 0.5 kg at rest on a smooth horizontal table is attached to two points p and q , which are 1.2 m apart, by two light elastic strings. The string attached to p has a natural length 0.4 m and elastic constant 75 N/m. The string attached to q has a natural length 0.6 m and elastic constant 50 N/m. (i) Find the equilibrium position. (ii) Prove that if the particle is displaced in the direction , through such a distance that neither string goes slack and is then released, it moves with simple harmonic motion.
a , b are two points 4 m apart on a smooth horizontal table and n is the midpoint of [ ab ]. A particle of mass 0.1 kg is held at n by two elastic strings the other ends of which are attached to a , b respectively. Each of the strings is of natural length 1 m and elastic constant 5 N/m. If the particle is then drawn aside along ab and released from rest when 1.5 m from b , show that it moves with simple harmonic motion. Find the period, and the least time taken to reach a point 1.75 m from b.
1990 (b) The depth of water in a harbour is assumed to rise and fall with time in simple harmonic motion. On a certain day the low tide had a height of 13 m at 12:58 p.m. and the following high tide had a height of 18 m at 6:58 p.m. If a ship requires a depth of 16. 5 m of water before it can leave the harbour, find the latest time on that day that the ship can leave the harbour.
Guide to answering the Exam Questions 2009 (a) (i) Would be straightforward if not for the ε term. Either use your knowledge of differentiation or else just learn this off because it does come up o am go h-am) (ii) Straightforward if you choose the correct equation. Answer: v = 3.03 cm/s
2009 (b) This was quite simply an absolutely horrendous question (and it doesn’t help when letters are used instead of numbers). You get an expression for the maximum friction force (fair enough; F = μR) and a separate expression for the motion involving the “N oscillations per minute” information. I can see where the value for ω comes from, but I would have thought that the F = mrω^2 expression referred to circulation motion and not SHM. Presumably I would be wrong. Once you get an expression for both these terms simply equate to get r = amplitude = (893.65 μ)/N^2
2008 (a) (i) Straightforward. Ans: a = - 100 x (ii) Straightforward. Ans: a = 10.8 m s-
2007 (a) (i) Straightforward. Ans: a = - √(k/m) x (ii) Straightforward. Ans: T = 2π √(d/g)
2006 (a) (i) Straightforward. Ans: amax = 3.78 m s-2. (ii) Straightforward. Work out the time it takes to go from equilibrium to the point 4 metres out, then take this away from T/4 (the time it takes to go from equilibrium to the extreme) and multiply this by two to get the time to go out and back in. Ans: t = 3.24 seconds.
2005 (b) Nice question. Find ω as normal. The time to reach o can then be broken into two stages: time to travel a distance a from the release point. The string is stretched for all of this so we need to find the time (which will be T/4) and use SHM formula to find the velocity at the end of this stage. From here on the string is no longer taut, so the particle is simply moving in a horizontal direction with whatever velocity it had when it first got to that point.
2004 (b) (i) Straightforward, but only if you had seen it before. x = a cos (ω t − β), d x /dt = - a ω sin (ω t − β), d^2 x /dt^2 = a ω^2 cos (ω t − β ). Can you say why? (ii) First find ω, then get two equations with A and β, then solve. Ans: ω = π/8, β = 1.176 radians, A = 13
2003 (a) (i) Straightforward. Ans: A = 4 cm, (ii) Straightforward. Ans: v = 2√7 cm/sec.
2001 (a) (i) Straightforward. Ans: A = 4.16, T = 3. (ii) Straightforward. Ans: vmax = 2√13 cm/sec.
2001 (b) (i) Straightforward. Distance from p = 5l/4. (ii) Time taken is in two stages; the first is from the release point to equilibrium position which is equivalent to T/4. The second stage is from equilibrium position to a distance of l/4 above equilibrium position (above this point the string will be slack. Ans: T = π/3√(l/g)
2000 (b) This is a tricky SHM question, but at least it is broken up into parts. (i) Straightforward. x1 = (k + 0.3g)/2k and x2 = (k - 0.3g)/2k. (ii) Slightly tricky. If the lower part of the string is taut then there must be an extension, i.e. x2 > 0. Use this expression to get a minimum value for k as 0.3g. (iii)Straightforward, if testing due to the expressions being slightly complex. T = π√(0.6/k)
1999 (a) A little tricky in that there is an ε term in there because the particle is initially 8 cm from the centre but if you’ve dealt with this before then it should be fine. Answer: x = √66 sin(4t + 1.4)
1999 (b) (i) Straightforward; equilibrium position is 0.48 m from p. (ii) Straightforward.
1998 (a) Very unusual (and therefore imo rather nasty) to get asked a definition, but physics students should know it off by heart at least. This is one of those questions which is very easy if you have seen it before, but if you haven’t... To show that the particle is describing SHM, simply treat the two components (7sin ωt and 24 cos ωt) separately and proceed as normal. To calculate the amplitude simply look at the amplitude of both components (7 and 24) and get the magnitude using Pythagoras. Answer: Amplitude = 25
1998 (b) Straightforward vertical SHM question. Note that usually you are given k and need to calculate equilibrium position, but in this case you are actually given equilibrium position and need to calculate k. There are two stages involved here;
1997 (a) No. 7 First up, it was unusual to find a SHM question as part (a) on No.7, which is always statics, but there you go. As a SHM question it was fine. (i) This was very straightforward and remarkably similar to No. 6 part (i) of the same year. No SHM for this part. Forces up = forces down and forces left = forces right. You also needed a little trigonometry to figure out the angle at the bottom. Answer: T = 13W/ (ii) Straightforward. Answer: k = 125W/
