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**DEFINITIONS TEST!! You have 12 minutes!**

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 = -ω2x) Solve problems using a = -ω2x Apply the equations x = x0cosωt, x = x0sinωt, v = v0sinωt, v = v0cosωt, and v = ±ω√(x02 – x2)

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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.**

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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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**f = 1/T T = 1/f Period and frequency**

Period and frequency are reciprocals of each other f = 1/T T = 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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Hooke’s 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 displacement**

Time displacement

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**Graph of motion A graph of the motion will have this form Amplitude x0**

Period T Time displacement

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**Graph of motion Notice the similarity with a sine curve 2π radians**

angle π/2 π 3π/2 2π

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**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

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Graph of motion Amplitude x0 Period T Time displacement

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**Graph of motion x = x0sinωt**

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

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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)

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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)

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**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)

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**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)

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**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)

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Maximum velocity?

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

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**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)

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**Let’s 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 = -ω2x a = -(2.5)20.5 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 = -ω2x x = -a/ω2 = 8/(8π)2 = 1/8π2 = 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, 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?

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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, 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 = 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, 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 = m.s-2

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**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

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**Let’s try some questions!**

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

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Simple Harmonic Motion

Simple Harmonic Motion

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