Simple Harmonic Motion The mathematical descriptions of oscillations

Objectives Show that motion is periodic Prove that 𝑣 𝑚𝑎𝑥=2𝜋𝑓𝐴 Define SHM and prove the equation 𝑎=− 2𝜋𝑓 2𝑥

Free Oscillations These are ones where we can ignore frictional forces. In this case an oscillator would continue to oscillate indefinitely. This is reasonable for a small number of some oscillations

Examples Pendulum Child on swing Mass on a spring

Position-time The position (x) of a simple harmonic oscillator is sinusoidal. Where we start measuring from is arbitrary.

Position-time 𝑥 𝑡 =𝐴𝑠𝑖𝑛(2𝜋𝑓𝑡) A is the amplitude f is the frequency t is the time

Definitions T = time period This is the time for one complete oscillation i.e. to go from ABCBA

f = frequency The number of complete oscillations in unit time Frequency and time are related by 𝑓= 1 𝑇

Angular frequency 𝜔= 2𝜋 𝑇 𝜔=2𝜋𝑓 →𝑓= 𝜔 2𝜋 So angular frequency is the number of complete oscillations per unit time.

Phase difference φ One complete sine wave is 2π radians Phase difference φ One complete sine wave is 2π radians. One wave lagging behind another by half a complete cycle has phase difference of π.

Phase difference of 𝜋 2

If we include phase difference 𝑥=𝐴 sin (2𝜋𝑓𝑡 + φ)

Example An oscillator with time period of 2.0s and amplitude 10 cm is oscillating. Calculate its displacement at t = 0.5s t = 1.0s t = 1.5s t = 2.0s

Velocity The displacement- time graph is sinusoidal as we have seen.

Velocity is the gradient of the displacement-time graph 𝑥(𝑡)=𝐴𝑠𝑖𝑛(2𝜋𝑓𝑡) If we differentiate 𝑣(𝑡)= 𝑑𝑥 𝑑𝑡 = 𝑑 𝑑𝑡 𝐴𝑠𝑖𝑛(2𝜋𝑓𝑡) 𝑣(t)=2𝜋fA𝑐𝑜𝑠(2𝜋𝑓𝑡)

Maximum speed in SHM 𝑣(t)=2𝜋fA𝑐𝑜𝑠(2𝜋𝑓𝑡) This function is maximised when 𝑐𝑜𝑠(2𝜋𝑓𝑡) = 1. So it follows that 𝑣𝑚𝑎𝑥=2π𝑓𝐴

Acceleration 𝑎= 𝑑𝑣 𝑑𝑡 𝑎= 𝑑 𝑑𝑡 𝐴2𝜋𝑓𝑐𝑜𝑠(2𝜋𝑓𝑡) 𝑎=−𝐴4𝜋𝑓2sin(2𝜋𝑓𝑡)

Acceleration This is best written as 𝑎=−(-2πf)2 Asin(2𝜋𝑓𝑡)

Acceleration Or better still 𝑎=− 2𝜋𝑓 2𝑥

Simple Harmonic Motion defined From the equation: 𝑎=−(2𝜋𝑓)2𝑥 The definition of SHM is: Acceleration is proportional to displacement Acceleration is directed to the centre

Example An oscillator moves in simple harmonic motion with time period 1.5 s and amplitude 2cm. Calculate: The frequency. The maximum acceleration. State where this occurs. The maximum speed. State where this occurs.

Example A platform is made to oscillate with amplitude 5 cm but ever increasing frequency. A small package is placed on the platform. Calculate the maximum frequency of oscillation to keep the package from losing contact with the platform.

Example An oscillator has amplitude of 10cm and period 0.25s. Calculate: Maximum velocity Maximum acceleration Sketch individual graphs of displacement, velocity and acceleration against time. State the position(s) where v and a are maximums.

Mass-Spring System Consider a mass of mass m attached to a spring of constant k. The tension in the spring is then given by Hooke’s law: 𝐹=−𝑘𝑥 The – sign indicates that the force is directed to the centre.

Mass-Spring System Since we know that 𝐹=𝑚𝑎 It also follows that 𝑚𝑎=−𝑘𝑥 Or, rearranging 𝑎= −𝑘𝑥 𝑚

Mass-Spring System So we can write −𝑘𝑥 𝑚 =−(2𝜋𝑓)2𝑥 𝑘 𝑚 = 4π2f2 Use this to get an equation for the time period T.

Mass-Spring system 𝑇=2𝜋 𝑚 𝑘

Simple Pendulum

Simple Pendulum If we examine the forces then the force towards the mid-point is mgsinθ

Simple Pendulum F = ma =mgsinθ The arc length s = Lθ If θ is small then we use the small angle approximation….

Continued.. ma = mgsinθ Since sinθ≈θ ma = mgθ = mgs/L From Hooke’s law we had mg =-kx mg = -kL, replacing x with L.

k =mg/L 𝑇=2𝜋 𝑚 𝑘 𝑇=2π 𝑚 𝑚𝑔/𝐿 𝑇=2π 𝐿 𝑔

Time period of pendulum with small angle 𝑇=2π 𝐿 𝑔

Experiment Set pendulum in motion to find its time period T. Alter the length L and repeat until 6-8 sets of readings obtained. Plot T2 vs L to find the acceleration due to gravity.

Energy in SHM Consider a simple pendulum oscillating back and forth. If we ignore air resistance then GPE KE  GPEKE etc etc. That is to say

Energy in SHM …total energy of the system is constant. And the amount of KE and GPE varies with position.

Energy in SHM

Example A mass 0.7 kg is attached to a spring of k = 50 Nm-1. The mass is pulled down by 10 cm and then released, executing SHM. Calculate: The energy stored in the spring. The maximum speed of the mass and where this occurs. The maximum acceleration of the mass.

Answer 𝐸= 1 2 𝑘𝑥2 = 0.25 J 𝐸𝑘= 1 2 𝑚𝑣2 The energy stored in the spring is transferred to KE  v = 0.85 ms-1 . As ever, this occurs at midpoint. From F = -kx, the force acting on the mass is 5N so its acceleration is , 𝑎= 𝐹 𝑚 = 7.14 ms-2