Upload
mackenzie-connolly
View
224
Download
1
Embed Size (px)
Citation preview
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.
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)