DEFINITIONS TEST!! You have 12 minutes! Do now! DEFINITIONS TEST!! You have 12 minutes!

Topic 4 Oscillations and Waves

Aims Remember the terms displacement, amplitude, frequency, period and phase difference. Define simple harmonic motion (a = -ω2x) Solve problems using a = -ω2x Apply the equations x = x0cosωt, x = x0sinωt, v = v0sinωt, v = v0cosωt, and v = ±ω√(x02 – x2)

Displacement - x The distance and direction from the equilibrium position. = displacement

Amplitude - A The maximum displacement from the equilibrium position.

Period - T The time taken (in seconds) for one complete oscillation. It is also the time taken for a complete wave to pass a given point. One complete wave

Frequency - f The number of oscillations in one second. Measured in Hertz. 50 Hz = 50 vibrations/waves/oscillations in one second.

f = 1/T T = 1/f Period and frequency Period and frequency are reciprocals of each other f = 1/T T = 1/f

Phase difference is the time difference or phase angle by which one wave/oscillation leads or lags another. 180° or π radians

Phase difference is the time difference or phase angle by which one wave/oscillation leads or lags another. 90° or π/2 radians

Simple harmonic motion (SHM) periodic motion in which the restoring force is proportional and in the opposite direction to the displacement

Hooke’s law What can you remember?

Simple harmonic motion (SHM) periodic motion in which the restoring force is in the opposite durection and proportional to the displacement F = -kx

Graph of motion A graph of the motion will have this form displacement Time displacement

Graph of motion A graph of the motion will have this form Amplitude x0 Period T Time displacement

Graph of motion Notice the similarity with a sine curve 2π radians angle π/2 π 3π/2 2π

Graph of motion x = x0sinθ Notice the similarity with a sine curve Amplitude x0 x = x0sinθ 2π radians angle π/2 π 3π/2 2π

Graph of motion Amplitude x0 Period T Time displacement

Graph of motion x = x0sinωt Amplitude x0 Period T x = x0sinωt Time displacement where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

When x = 0 at t = 0 Amplitude x0 Period T x = x0sinωt Time displacement where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

When x = x0 at t = 0 Amplitude x0 Period T x = x0cosωt displacement Time where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

When x = 0 at t = 0 x = x0sinωt v = v0cosωt Amplitude x0 Period T Time displacement where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

When x = x0 at t = 0 x = x0cosωt v = -v0sinωt Amplitude x0 Period T displacement Time where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

To summarise! When x = 0 at t = 0 x = x0sinωt and v = v0cosωt When x = x0 at t = 0 x = x0cosωt and v = -v0sinωt It can also be shown that v = ±ω√(x02 – x2) and a = -ω2x where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

Maximum velocity?

Maximum velocity? When x = 0 At this point the acceleration is zero (no resultant force at the equilibrium position).

Maximum acceleration?

Maximum acceleration? When x = +/– x0 Here the velocity is zero

Oscillating spring We know that F = -kx and that for SHM, a = -ω2x (so F = -mω2x) So -kx = -mω2x k = mω2 ω = √(k/m) Remembering that ω = 2π/T T = 2π√(m/k)

Let’s do a simple practical! T = 2π√(m/k)

We need to try some examples!

Example 1 Find the acceleration of a system oscillating with SHM where ω = 2.5 rad s-1 and x = 0.5 m to 2.s.f.

Example 1 Find the acceleration of a system oscillating with SHM where ω = 2.5 rad s-1 and x = 0.5 m to 2.s.f. Using a = -ω2x a = -(2.5)20.5 a = -3.13 m.s-2

Example 2 Find the displacement at a point to 2.s.f. of a system of frequency 4 Hz when its acceleration is -8 ms-2.

Example 2 Find the displacement at a point to 2.s.f. of a system of frequency 4 Hz when its acceleration is -8 ms-2. ω = 2πf = 2π x 4 = 8π a = -ω2x x = -a/ω2 = 8/(8π)2 = 1/8π2 = 0.013 m

Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s-1, xo = 0.5 m and t = 1.0 s. What is the maximum acceleration of this system to 3.s.f? What is the maximum velocity of this system?

Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s-1, xo = 0.5 m and t = 1.0 s. What is the maximum acceleration of this system to 3.s.f? What is the maximum velocity of this system? x = xosinωt (when x = 0 at t = 0) x = 0.5sin(5.0 x 1.0) = 0.5sin5 = -0.479 m

Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s-1, xo = 0.5 m and t = 1.0 s. What is the maximum acceleration of this system to 3.s.f? What is the maximum velocity of this system? a = -ω2x Maximum acceleration when x = ±xo amax = -ω2xo = -(5)2 x 0.5 = -12.5 m.s-2

Example 3 v = ±ω√(x02 – x2) Maximum velocity when x = 0 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s-1, xo = 0.5 m and t = 1.0 s. What is the maximum acceleration of this system to 3.s.f? What is the maximum velocity of this system? v = ±ω√(x02 – x2) Maximum velocity when x = 0 vmax = ± ω√(x02 – x2) = ±5.0√(0.5)2 = ±2.5 m.s-1

Let’s try some questions! Finish for homework. Due Thursday 26th February (two days before Mr Porter’s virtual birthday)