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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.

amplitude

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.

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

Time

displacement

Graph of motion

A graph of the motion will have this form

Time

displacement

Amplitude

x0

Period T

Graph of motion

Notice the similarity with a sine curve

angle

2π radians

π/2 π 3π/2 2π

Graph of motion

Notice the similarity with a sine curve

angle

2π radians

π/2 π 3π/2 2π

Amplitude

x0 x = x0sinθ

Graph of motion

Time

displacement

Amplitude

x0

Period T

Graph of motion

Time

displacement

Amplitude

x0

Period Tx = x0sinωt

where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

When x = 0 at t = 0

Time

displacement

Amplitude

x0

Period Tx = x0sinωt

where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

When x = x0 at t = 0

Time

displacement

Amplitude

x0

Period Tx = x0cosωt

where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

When x = 0 at t = 0

Time

displacement

Amplitude

x0

Period T

x = x0sinωt v = v0cosωt

where ω = 2π/T = 2πf = (angular frequency in rad.s-1)

When x = x0 at t = 0

Time

displacement

Amplitude

x0

Period T

x = x0cosωt

v = -v0sinωt

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 = -mω2xk = 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

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)