Oscillations

1. Oscillations • 14. Oscillations • Content • 14.1 Simple harmonic motion • 14.2 Energy in simple harmonic motion • 14.3 Damped and forced oscillations: resonance • Learning Outcomes • (a) describe simple examples of free oscillations. • * (b) investigate the motion of an oscillator using experimental and graphical methods. • (c) understand and use the terms amplitude, period, frequency, angular frequency and phase difference and express the period in terms of both frequency and angular frequency. • (d) recognise and use the equation a = – ω2x as the defining equation of simple harmonic motion.

2. (e) recall and use x = xo sin ωt as a solution to the equation a = – ω2x. • (f) recognise and use v = vo cos ωt, v = ± ω√ (x2o − x2) • * (g) describe with graphical illustrations, the changes in displacement, velocity and acceleration during simple harmonic motion. • (h) describe the interchange between kinetic and potential energy during simple harmonic motion. • * (i) describe practical examples of damped oscillations with particular reference to the effects of the degree of damping and the importance of critical damping in cases such as a car suspension system. • (j) describe practical examples of forced oscillations and resonance. • * (k) describe graphically how the amplitude of a forced oscillation changes with frequency near to the natural frequency of the system, and understand qualitatively the factors which determine the frequency response and sharpness of the resonance. • (l) show an appreciation that there are some circumstances in which resonance is useful and other circumstances in which resonance should be avoided.

3. Oscillations and vibrations • Vibrations and oscillations occur all the time and are everywhere. • Vibrations are physical evidence of waves, such as a loud stereo shaking a table, i.e. sound waves cause vibrations • One complete movement from the starting point or rest point or equilibrium position and back to the starting point or rest position or equilibrium position is known as an oscillation • The time taken for one complete oscillation is referred to as the period T of the oscillation • The number of oscillations per unit time is the frequency f • Frequency f = 1/T , may be measured in hertz (1 Hertz = 1 s-1) or in min-1, hour-1 etc • The distance from the equilibrium position is known as the displacement and it is a vector quantity since the displacement may be on either side of the equilibrium position • The amplitude (a scalar quantity) is the maximum displacement

4. Examples of oscillatory motion • beating of a heart • a simple pendulum • a vibrating guitar string • vibrating tuning fork • atoms in solids • air molecules oscillate when sound waves travel through air. • oscillations in electromagnetic waves such as light and radio waves • oscillations in alternating current and voltage.

5. Recap from study of waves • Some oscillations maintain a constant period even when the amplitude of the oscillation changes. This is known as isochronous and has been made use of in timing devices • Galelli Galileo discovered this for a pendulum. A pendulum swinging with a large amplitude is not isochronous

6. Displacement-time graphs • It is possible to plot displacement-time graphs for oscillators • The graph describing the variation of displacement with time may have different shapes depending on the oscillating system • For many oscillators the displacement-time graph of a free oscillation is approximately a sine or cosine curve

7. Simple harmonic motion (shm) • A sinusoidal displacement time graph is a characteristic of an important type of oscillation called simple harmonic motion(shm) • In harmonic oscillators the amplitude is constant with time • SHM is defined as the motion of a particle about a fixed point such that its force F or acceleration a is proportional to its displacement x from the fixed point, and is directed towards the point • F is known as the restoring force • Mathematically it is defined as a = - ω2x where ω is the angular frequency and is equal to 2πf

8. cont… • The defining equation is represented in a graph of a against x as a straight line of negative gradient through the origin. • Gradient is negative because of the minus sign in the equation which represents that acceleration is always directed towards the fixed point from which the displacement is measured • This means that in shm, acceleration is directly proportional to the displacement/distance from the fixed point and is always directed to that point • Acceleration is always opposite to the displacement since the force is also opposite to the displacement a 0 x

9. Comparisons • In linear motion, acceleration is constant in magnitude and direction • In circular motion acceleration is constant in magnitude but not direction • In simple harmonic motion the acceleration changes periodically in magnitude and direction

10. Solution of equation for shm • In order to find the displacement time relation for a particle moving in shm, we need to solve the equation a = - ω2x which requires mathematics beyond the requirements of A/AS • However we need to know the form of the solution x = x0 sin ωt or x = x0 cos ωt where x0 is the amplitude of the oscillation • The solution x = x0 sin ωt is used when at time t = 0, the particle is at its equilibrium position where x = 0, and conversely if at time t = 0 the particle is at its maximum displacement, x = x0 the solution is x = x0 cos ωt

11. Velocity & acceleration for shm • The velocityv of the particle is given by the expressions v = x0ωcosωt when x = x0 sin ωt v = -x0ω sin ωt when x = x0cosωt • The maximum speed is given by v0 = x0ω • An alternate expression for the velocity is v = ±ω√(x02 – x2) (which will be derived next) • The accelerationa of the particle is given by the expressions a = -x0ω2 sin ωt when x = x0 sin ωt a = -x0ω2 cosωt when x = x0cosωt

12. Displacement, velocity and acceleration graphs x t v t a t Displacement (x), velocity (v) & acceleration time graph

13. Alternate expression for velocity • Recall that x = x0 sin ωt and v = x0ω cos ωt • So sin ωt = x/x0and cos ωt = v/(x0ω) • Trigonometric relationship between sine and cosine is sin2θ + cos2θ = 1 • Applying the above relationship, we have x2/x02 + v2/(x02ω2) = 1 which gives v2 = x02ω2 - x2 ω2 , hence v = ± ω√(x02 - x2)

14. Example The displacement xat time t of a particle moving in shm is given by x = 0.25 cos 7.5t where x is in metres and t is in seconds. a) use the equation to find the amplitude, frequency and period for the motion b) find the displacement when t = 0.50 s Solution a) Compare the equation with x = x0cosωt The amplitude x0 = 0.25 m, ω = 2πf = 7.5 rad/s, therefore f = 1.2 Hz and period T = 1/f = 0.84 s b) Substitute t = 0.50 s in the equation ωt = 7.5 x 0.50 = 3.75 rad = 215° so x = 0.25 cos 215° = -0.20 m

15. Worked examples of shm • Mass on a helical spring • Simple pendulum

16. Mass on a helical spring – Hooke’s law • Consider a mass m suspended from a spring • The weight mg is balanced by the tension T in the spring • When the spring is extended downwards by an amount x away from the equilibrium position, there is an additional upward force called the restoring force in the spring given by F = - kx • When the mass is released the restoring force F pulls the mass upwards towards the equilibrium position. The minus sign shows the direction of this force. • As the force is proportional to the displacement, the acceleration is also proportional to the displacement and is directed towards the equilibrium position meeting the condition for shm • The full theory shows that the period of oscillation T = 2π√(m/k) • since F = ma, then ma = - kx hence a = - (k/m)x = dv/dt = d2x/dt2

17. The simple pendulum • A simple pendulum is a point mass m on a light inelastic string although in real experiments we use a finite pendulum bob of finite mass • When the bob is pulled aside through an angle and released, there will be a restoring force acting in the direction of the equilibrium position • Because the pendulum moves in an arc of a circle, the displacement will be an angular displacement rather than a linear displacement • The 2 forces on the bob are its weight mg and the tension T in the string • The component of the weight along the direction of the string mg cos θ, is equal to the tension T in the string • The component of the weight at right angles to the direction of the string, mg sin θ , is the restoring force F. This makes the bob accelerate towards the equilibrium position • The restoring force depends on θ . As θincreases the restoring force is not proportional to the displacement and so the motion is oscillatory but not shm, but if the angle is kept small (less than 5°), θ is proportional to sin θ and exhibits shm (check using your calc) • The full theory shows period of oscillation T = 2π√(l/g) where g is the acceleration of free fall • A simple pendulum can be used to experimentally determine g by repeating the experiment with different lengths of pendulum and plotting a graph of T2 against 4π2/g

18. Hooke’s law & shm • Any system which obeys Hooke's Law exhibits shm but i) extensions must not exceed the limit of proportionality ii) the spring must have small oscillations as large amplitude oscillations may cause the spring to become slack iii) the spring should have no mass; if the mass is > 20x the mass of the spring, the error is 1% • This example of shm is a particularly useful model for interatomic forces and vibration of molecules containing atoms oscillating as if connected by tiny springs • The frequency of oscillation can be measured using spectroscopy which gives direct information about the bonding

19. Example A light spring of spring constant k hangs vertically from a fixed point and a mass m is attached to its free end. a) State 2 conditions that must be met before the subsequent motion may be considered to be simple harmonic. b) Derive an expression for the period T of the resulting motion. Solution a) 2 conditions for shm are: a) The equilibrium position due to the mass is within the Hooke’s law limit of the spring b) the mass is given a small vertical displacement such that the spring’s Hooke’s law limit is not exceeded b) Let x = displacement of mass m, a = acceleration of mass m, F = ma = -kx Force in a spring is, - kx = ma , hence a = - (k/m)x As a is proportional to - x , so resulting motion is shm i.e. a = - ω2x a = - (k/m)x = - ω2x, so angular frequency ω = √(k/m) Therefore period T = 2π/ω = 2π√(m/k)

20. Example A light string of length l hangs vertically from a fixed support and a mass m is attached to its free end. The mass is given a horizontal displacement and released to swing freely. a) State a condition which must be satisfied before the resulting oscillation may be considered shm. b) Derive an expression for the period T of the resulting motion. Solution θ a) A required condition is that the angular displacement θ is small l b) Let x = displacement of mass m, a = acceleration of mass m In the direction perpendicular to string, F = ma - mg sin θ = ma, so - g sin θ = a For small θ , sin θ ≈ x/l, so - gx/l ≈ a x As a is proportional to - x , so resulting motion is shm i.e. a = - ω2x Hence, - (g/l)x = - ω2x, so ω = √(g/l) mg Therefore period T = 2π/ω = 2π√(l/g)

21. Example A helical spring is clamped at one end and hangs vertically. It extends by 10 cm when a mass of 50 g is hung from its free end. Calculate: a) the spring constant of the spring b) the period of small amplitude oscillations of the mass Solution a) k = F/x, k = 4.9 Nm-1 b) T = 2π√(m/k) T = 0.63 s

22. Energy changes in shm • A system exhibiting simple harmonic motion would possess a constant total energy at all points of time • The total energy normally comprises a portion of potential energy and another balanced portion of kinetic energy. • There is thus a continuous interchange of the two energies during oscillations. • For example, a weighted helical spring has a total energy that is the sum of the kinetic energy of the moving mass and the stored elastic potential energy of the spring. • Plotting on the same graph for energy versus time/displacement, the two sinusoidal curves are completely out of phase. • It can be proven that the total energy of a weighted spring is ½ mω2xo2 which is a constant. .

23. Energy vs time graph K.E. P.E. total energy energy 0 0 T/4 T/2 3T/4 T time Energy versus time graph

24. Displacement, velocity and acceleration graphs x t v t a t Displacement (x), velocity (v) & acceleration time graph

25. Energy changes in shm • The kinetic energy of a particle of mass m oscillating with shm is ½mv2 and from the earlier derivation v2 = x02ω2 - x2 ω2 • So k.e Ek at displacement x is ½mω2(x02- x2) • To find the potential energy Epwe need to find the work done against the restoring force; since F = ma , Fres = - mω2x but average restoring force = ½mω2x • Hence work done = average restoring force x displacement = ½mω2x2 • The total energy Etotof the oscillating system is given by • Etot= Ek +Ep = ½mω2(x02- x2) + ½mω2x2 = ½mω2x02 • This total energy is constant as it merely expresses the law of conservation of energy • Pg 272 Chris Mee figs 10.22, 10.23,10.24

26. Example A particle of mass 60 g oscillates with shm with angular frequency of 6.3 rad/s and amplitude 15 mm. Calculate a) the total energy b) the k.e and p.e at half amplitude (i.e. at x = 7.5 mm) Solution Etot= Ek+Ep = ½mω2(x02- x2) + ½mω2x2 = ½mω2x02 a) Etot = ½mω2x02= 2.7 x 10-4 J b) Ek = ½mω2(x02- x2) = 2.0 x 10-4 J Ep= ½mω2x2 = 0.7 x 10-4 J

27. Natural frequency & resonance • A particle is said to be undergoing free oscillations when the only external force acting on it is the restoring force • The total energy remains constant at all points of time • A free oscillation is one where an object or system oscillates in the absence of any damping forces, and it is said to oscillating in its natural frequency • In real situations, frictional and other resistive forces cause the oscillator’s energy to be dissipated, and this energy is converted eventually into heat energy. The oscillations are said to be damped • When one object vibrates at the same frequency as another it is said to be in resonance • The ‘swing’ of a frictionless pendulum is an example of a free oscillation.

28. Resonance • In the absence of external forces to an oscillating system, the system oscillates at its natural frequency f0. The only forces acting are the internal forces of the oscillating system • When an external force is applied to an oscillating system, the system is under forced oscillations and will vibrate at the frequency of the applied force rather than at the natural frequency of the system • Whether or not the forcing frequency equals the natural frequency, the oscillations are said to be forced when a periodic force acts. • When the forcing frequency is equal to the natural frequency, netenergy is taken in and the amplitude of oscillation builds up further and the applied periodic force is said to have set the system in resonance. Under such condition, further resonance will result in more energy being taken in to build up the amplitude further. • Resonance occurs when a system is forced to oscillate at its natural frequency by the driving frequency • When resonance occurs, the amplitude of the resulting oscillations is a maximum as maximum energy is transferred from the forcing system E.g. Barton's pendulum – only the pendulum with the same length as the original will oscillate with the biggest amplitude • Applications – wind instruments, excessive noise from a moving bus, radio & tv tuning • The Tacoma Narrows suspension bridge in Washington State, USA in 1940 collapsed due to a moderate gale (of same frequency as natural frequency of bridge) setting the bridge into resonance until the main span broke up

29. Damped oscillations • A damped oscillationis one where frictional forces present gradually slow down the oscillation and the amplitude decreases with time i.e. decreasing energy • Damped oscillations are divided into under-damped, critically damped and over-damped oscillations • An under-damped(lightly damped) oscillation is one where the amplitude of oscillation or displacement of the system decreases with time. Example: oscillation of a simple pendulum with the damping or dissipative force as air resistance • In a critically damped system, oscillations are reduced to zero in the shortest possible time. Examples: moving coil ammeter or volt meter, shock absorber, door closer • In an over-damped(heavy damping) system, a displacement from its equilibrium position takes a long time for the displacement to be reduced to zero. Example: door dampers

30. Damped oscillations Displacement vs Time Graph x over-damped t under-damped critically-damped

31. Effects of damping on forced oscillations • Pg 277 Chris Mee fig 10.29 & 10.30 • As the degree of damping increases: • The amplitude of oscillation at all frequencies is reduced • The frequency at max amplitude shifts gradually towards lower frequencies • The peak becomes flatter

32. Electrical resonance Electrical oscillators made from combinations of capacitors and inductors(coils) can also be forced into oscillations or be made to resonate This is the basis of tuning in electronic circuits which pick out the required transmission in a receiver The natural frequency of an electrical oscillator depends on the capacitor and inductance of the coil used. By varying the capacitance, we can tune in to different ‘channels’ The range of frequencies selected depends on the damping which in turn depends on the resistance in the circuit 32