Topic 4 – Oscillations and Waves4.1 Simple Harmonic Motion
Oscillations● There are many systems, both natural and manmade, that vibrate back and forth around anequilibrium point.● These systems are said to regularly oscillate.● Common examples are:● A mass on a spring● A pendulum● Electrons under alternating current
Key Terms● The equilibrium point is that point where thesystem will naturally rest.● e.g. for a pendulum – bottom centre● For a mass on a spring – the point where theupwards pull of the spring equals the downward pullof the weight.● The displacement (x) of the system is thevector displacement of the system from itsequilibrium point.● Usually the displacement is considered in 1dimension and is given the symbol x even if thedisplacement is vertical.
Key Terms● The amplitude (A) of an oscillation is themaximum displacement of the system.● It is the height of a wave from its equilibrium point.● It is half the peak to trough height.● The wavelength (λ) of a moving wave is thedistance from peak to peak in the spacedimension.
Key Terms● The time period (T) is the time taken in seconds tocomplete 1 complete cycle.● This is the time from peak to peak in the timedimension.● A cycle is complete when the system is back in itsinitial state.● e.g. for a pendulum, when the bob is at its lowest pointand travelling in the same direction as at the start.● The frequency (f) of the system is the number ofoscillations per second.● It is the inverse of the time period.● Frequency is measured in Hz or s-1
Key Terms● A sine wave has a period of 2π radians and atime period of T seconds.● Therefore its angular displacement (on an x-θgraph) at any time is:● The angular frequency (ω) in rad s-1istherefore:θ=2πTtω=2πT=2π f
Questions● Calculate the frequency and angular frequencyof:● A pendulum of period 4s● A water wave of period 12s● Mains electricity of period 0.02s● Laser light with period 1.5 fs
Key Terms● A sinusoidal wavehas is an oscillationwith the followingproperties.● It has an amplitude of1.● It has a period of 2πradians● It has an initialdisplacement of +0.x0 π/2 π 3π/2 2πθx=sinθ
Key Terms● A cosine wave isidentical to a sinewave excepting that ithas an initialdisplacement of +1● It can be said that acosine wave is a sinewave with a phasedifference (Ф) of-π/2x0 π/2 π 3π/2 2πθx=cosθx=sin(θ−π2)
Oscillating Systems● An oscillating system is defined as one thatobeys the general equation:● Here● the amplitude is x0● The angular frequency ensures that the real timeperiod coincides with the angular period of 2πradians● The phase allows for an oscillation that starts at anypoint.x=x0 sin(ωt+ ϕ)
Oscillating Systems● If the oscillations begin at the equilibrium pointwhere displacement is zero at the start then:● If the oscillations begin at the end point wheredisplacement is a maximum at the start then:● This second form is more useful in moresituationsx=x0 sin(ωt)x=x0 cos(ωt)
Questions● A simple harmonic motion is initiated byreleasing a mass from its maximumdisplacement. It has period 2.00s andamplitude 16.0cm. Calculate the displacementat the following times:● t=0s● t=0.25s● t=0.50s● t=1.00s
Time x v0 0 +v0T/4x0 0T/2 0 -v03T/4-x0 0T 0 +v0Oscillating Systems● The rate of change of displacement (the speed) isgiven by the gradient of the displacement curve.● Assuming that:● Then:● The speed is therefore:x=x0 sin(ωt)v=v0 cos(ωt)
Questions● A bored student holds one end of a flexible rulerand flicks it into an oscillation. The end of theruler moves a total distance of 8.0cm andmakes 28 full oscillations in 10s.● What are the amplitude and frequency of the motionof the end of the ruler?● Use the displacement equation to produce a tableof x and t for t=0,0.04,0.08,0.012,...,0.036● Draw a graph of x versus t● Find the maximum speed of the end of the ruler.
Oscillating Systems● The rate of change of speed (the acceleration)is given by the gradient of the speed curve.● Using similar logic:● This has a very similar form to the displacementequation therefore:● Note that the acceleration is:● In the opposite direction to the displacement,● Directly proportional to the displacement.a=−a0 sin(ωt)a=−a0x0x
The SHM Equation● Any system undergoing simple harmonicmotion obeys the relationship:● It can be shown using calculus or centripetalmotion that● Therefore:a∝−xa0=ω2x0a=−ω2x
The SHM Equations● For a system startingat equilibrium● The general SHM equation applies to all simpleoscillating systems.a=−ω2xx=x0 sin(ωt)v=ω x0 cos(ωt)a=−ω2x0 sin(ωt)● For a system starting atmaximum displacementx=x0 cos(ωt)v=−ω x0 sin(ωt)a=−ω2x0 cos(ωt)
The SHM Equations● One final equation can be formed by squaringthe speed equation.● Because sin2θ + cos2θ =1v=ω x0cos(ωt)v2=ω2x02cos2(ωt)v2=ω2x02(1−sin2(ωt))v2=ω2( x02−x02sin2(ωt))v2=ω2( x02−x2)v=±ω√x02−x2
Questions● A body oscillates with shm decribed by:● x=1.6cos3πt● What are the amplitude and period of themotion● At t=1.5s, calculate the displacement, velocityand acceleration.
Questions● The needle of a sewing machine moves up anddown with shm. If the total vertical motion ofthe needle is 12mm and it makes 30 stitches in7.0s calculate:● The period,● The amplitude,● The maximum speed of the needle tip● The maximum acceleration of the needle tip.
Energy Changes● An oscillating system is constantly experiencingenergy changes.● At the extremes of displacement, the potentialenergy is a maximum.● Gravitational potential for a pendulum, elasticpotential for a spring● At the equilibrium position, the kinetic energy isa maximum
Kinetic Energy● Remember that kinetic energy is given by:● Substituting● Givesv=±ω √ x02−x2v2=ω2( x02−x2)EK =12m ω2( x02−x2)EK =12m v2
Total Energy● Remember that at the equilibrium point ALL theenergy of the system is kinetic.● The total energy of the system ETis therefore:● Note that the total energy of the system isproportional to the amplitude squared.ET =12m ω2(x02−02)ET =12m ω2x02
Potential Energy● The law of conservation of energy requires thatthe total energy of an oscillating system be thesum of the potential and kinetic energies.●● Therefore12mω2x02=12mω2(x02−x2)+ EPEP=12m ω2x2ET =EK + EP
Questions● A pendulum of mass 250g is released from itsmaximum displacement and swings with shm.If the period is 4s and the amplitude of theswing is 30cm, calculate:● The frequency of the pendulum● The maximum speed of the pendulum● The total energy of the pendulum● The maximum height of the pendulum bob.● The energies of the pendulum at t=0.2s.