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

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Topic 4 Oscillations and Waves

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Aims Remember the terms displacement, amplitude, frequency, period and phase difference. Define simple harmonic motion (a = -ω 2 x) Solve problems using a = -ω 2 x Apply the equations x = x 0 cosωt, x = x 0 sinωt, v = v 0 sinωt, v = v 0 cosωt, and v = ±ω(x 0 2 – x 2 )

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

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Amplitude - A The maximum displacement from the equilibrium position. amplitude

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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

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

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

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

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

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

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Hookes law What can you remember?

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

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

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Graph of motion A graph of the motion will have this form Time displacement Amplitude x 0 Period T

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

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Graph of motion Notice the similarity with a sine curve angle 2π radians π/2 π 3π/2 2π Amplitude x 0 x = x 0 sinθ

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Graph of motion Time displacement Amplitude x 0 Period T

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Graph of motion Time displacement Amplitude x 0 Period T x = x 0 sinωt where ω = 2π/T = 2πf = (angular frequency in rad.s -1 )

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

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

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

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

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To summarise! When x = 0 at t = 0 x = x 0 sinωt and v = v 0 cosωt When x = x 0 at t = 0 x = x 0 cosωt and v = -v 0 sinωt It can also be shown that v = ±ω(x 0 2 – x 2 ) and a = -ω 2 x where ω = 2π/T = 2πf = (angular frequency in rad.s -1 )

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Maximum velocity?

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

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Maximum acceleration?

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When x = +/– x 0 Here the velocity is zero

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

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Lets do a simple practical! T = 2π(m/k)

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We need to try some examples!

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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.

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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 = -ω 2 x a = -(2.5) a = m.s -2

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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.

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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 = -ω 2 x x = -a/ω 2 = 8/(8π) 2 = 1/8π 2 = m

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Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s -1, x o = 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?

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Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s -1, x o = 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 = x o sinωt (when x = 0 at t = 0) x = 0.5sin(5.0 x 1.0) = 0.5sin5 = m

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Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s -1, x o = 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 = -ω 2 x Maximum acceleration when x = ±x o a max = -ω 2 x o = -(5) 2 x 0.5 = m.s -2

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Example 3 For a SHM system of x = 0 at t = 0, find x when ω = 5.0 rad s -1, x o = 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(x 0 2 – x 2 ) Maximum velocity when x = 0 v max = ± ω(x 0 2 – x 2 ) = ±5.0(0.5) 2 = ±2.5 m.s -1

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Lets try some questions! Finish for homework. Due Thursday 26 th February (two days before Mr Porters virtual birthday)

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