Oscillations

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03/07/2022 Oscillations JC Moller English Martyrs Sixth Form College

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Oscillations. JC Moller English Martyrs Sixth F orm College. Circular Motion. Circular Motion. 1) Is this car travelling at constant speed? 2) Is this car travelling at constant velocity?. Centripetal Acceleration. V a. Δ V. V b. - PowerPoint PPT Presentation

Transcript of Oscillations

Page 1: Oscillations

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Oscillations

JC Moller

English Martyrs Sixth Form College

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Circular Motion

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Circular Motion

1) Is this car travelling at constant speed?

2) Is this car travelling at constant velocity?

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Centripetal AccelerationIf the velocity is changing then it must be accelerating...

Va

Vb

ΔV

This change in velocity is towards the centre of the circle so the acceleration and is towards the centre if the circle –

“Centripetal Acceleration”

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RadiansTo further understand circular motion we need to use a different system for measuring angles:

Old method New method

r

s

Angle = 300 Angle = s/r rad

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RadiansCalculate the following angles in radians:

1)

3) 4)

2)

2cm

5mm

8m

6cm

1.5cm

15cm

2.05cm

50.24m

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Centripetal AccelerationConsider a circle:

r

θ

v1

v2

θv1

v2

Δv

If we assume θ is very small then v1 = v2 = v

Therefore θ = Δv/v

v

v

Also θ = vΔt/r

Therefore Δv/v = vΔt/r

Δv/Δt = v2/r

a = v2/r

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More Exciting EquationsFrom the last slide a = v2/r but F=ma so centripetal force F = mv2/r

The “angular speed” is the “angular distance” divided by time, or ω = θ/t

The total time period T for one revolution must therefore be the time taken to complete 2π revolutions, or ω = 2π/T

“Frequency” is how often something happens every second, so T = 1/f. Therefore ω = 2πf

For a whole circle, v = 2πr/T. However, T = 2π/ω. Therefore v = rω

Acceleration a = v2/r, therefore a = rω2

Finally, this must mean that F = mrω2

F = mv2/r

ω = θ/t

ω = 2π/T

ω = 2πf

v = rω

a = rω2

F = mrω2

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Example questions1) A disc spins twice per second. Calculate its angular

speed.

2) Estimate the angular speed of the Earth.

3) Rob spins a conker around his head using a 50cm long string. The conker has a mass of 0.1kg and he spins it with a velocity of 2ms-1. Calculate the centripetal force.

4) Calculate the velocity of a satellite moving with an angular speed of 7x10-5 rads-1 and at an altitude of 700km above the Earth (radius 6370km).

5) Tom drives his car in circles. If he drives with an angular speed of 1 rads-1 how many times will he make a complete turn in 10 seconds?

6) If the combined mass of Tom and his car is 1000kg calculate the centripetal force if his turning circle has a radius of 3m.

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Simple Harmonic Motion

Definition: simple harmonic motion is when acceleration is proportional to displacement and is always directed towards equilibrium.

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Simple Harmonic MotionConsider a pendulum bob:

Let’s draw a graph of displacement against time:

Displacement

Time

Equilibrium position “Sinusoidal”

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SHM Graphs

Time

Displacement

Velocity

Acceleration

Time

Time

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The Maths of SHM

Time

Displacement

As we’ve already seen, SHM graphs are “sinusoidal” in shape:

Therefore we can describe the motion mathematically as:

x = x0cosωt

v = -x0ωsinωt

a = -x0ω2cosωt

a = -ω2x

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The Maths of SHM

Definition: simple harmonic motion is when acceleration is proportional to displacement and is always directed towards equilibrium.

Recall our definition of SHM:

a = -ω2x

This agrees entirely with the maths: a

x

Important – remember ω = 2π/T

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SHM questionsa

x

5

2

1) Calculate the gradient of this graph

2) Use it to work out the value of ω

3) Use this to work out the time period for the oscillations

4) Howard sets up a pendulum and lets it swing 10 times. He records a time of 20 seconds for the 10 oscillations. Calculate the period and the angular speed ω.

5) The maximum displacement of the pendulum is 3cm. Sketch a graph of a against x and indicate the maximum acceleration.

a

x

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SHM Maximum Valuesx = x0cosωt

v = -x0ωsinωt

a = -x0ω2cosωt

Consider our three SHM equations:

Clearly, the maximum value that sinωt can take is 1, therefore:

amax = -ω2x0

xmax = x0 (obviously)

vmax = -x0ω (or max speed = ωx0)

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SHM periods: Two examples

For a pendulum the only thing that affects the period is the length of the string:

T = 2π lg

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SHM periods: Two examplesFor a spring there are two things that affect the period – the mass and the spring constant:

T = 2πmk

Where k is defined as “the force needed to extend the spring by a given number of metres” (units Nm-1):

F = -kΔx

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More questions1) Define simple harmonic motion.

2) A pendulum in a grandfather clock has a period of 1 second. How long is the pendulum?

3) Luke sets up a 200g mass on a spring and extends it beyond its equilibrium. He then releases it and enjoys watching it bounce up and down. If the period is 10s what is the spring constant?

4) Nick is envious of this and sets up another system with a spring constant of 0.1Nm-1. If the spring oscillates every 8 seconds how much mass did he use?

5) Simon sets up a pendulum and records the period as being 3 seconds. He then lengthens the pendulum by 1m and does the experiment again. What is the new period?

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SHM recap questions1) Define SHM and state “the golden SHM equation”

2) A body is performing SHM and is temporarily at rest at time t=0. Sketch graphs of its displacement, velocity and acceleration.

a

x

10

5

3) A body is performing SHM as shown on this graph. Calculate its angular speed and its time period T.

4) What is this body’s maximum speed?

5) A 1kg mass is attached to a spring of spring constant 10Nm-1. The mass is pulled down by 5cm and released. It performs SHM. Calculate the time period of this motion.

6) Describe the energy changes in this system as it bounces up and down.

7) Calculate the length of a pendulum if it oscillates with a period of 5s.

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SHM: Energy change

Equilibrium position Energy

Time

GPE

K.E.

Total

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Damping

Time

Displacement

Friction is a common cause of damping in oscillating systems. Notice that the period of oscillation does not change but the amplitude decreases each cycle.

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ResonanceResonance occurs when the frequency of

a driving system matches the natural frequency of the system it is driving.

Watch videoWatch video

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The Millennium Bridge

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Earthquake Dampers

The Taipei 101 skyscraper in Taiwan, featuring a giant steel pendulum of mass 728 tonnes.