Do now!

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Do now! Can you hand your SHM questions in please?

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Do now!. Can you hand your SHM questions in please?. This lesson. Considering energy changes during SHM Applying E k = ½mω 2 (x o 2 – x 2 ) E T = ½mω 2 x o 2 E p = ½mω 2 x 2 where ω = 2πf = 2π/T. S.H.M. Where is the kinetic energy maxiumum? - PowerPoint PPT Presentation

Transcript of Do now!

Page 1: Do now!

Do now!

Can you hand your SHM questions in please?

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This lesson

• Considering energy changes during SHM• Applying Ek = ½mω2(xo

2 – x2)

ET = ½mω2xo2

Ep = ½mω2x2

where ω = 2πf = 2π/T

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S.H.M.Where is the kinetic energy maxiumum?

Where is the potential energy maximum?

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It can be shown that….

Ek = ½mω2(xo2 – x2)

ET = ½mω2xo2

Ep = ½mω2x2

where ω = 2πf = 2π/T

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Example

A mass-spring system oscillates with an amplitude of 3.5cm.  If the force constant of the spring of 250Nm-1 and the mass is 0.5kg, determine(a) the total energy of the system,(b) the maximum speed of the mass(c) the maximum acceleration

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• A mass-spring system oscillates with an amplitude of 3.5cm.  If the force constant of the spring of 250Nm-1 and the mass is 0.5kg, determine(a) the total energy of the system,(b) the maximum speed of the mass(c) the maximum acceleration

(a) We have m = 0.5kg, xo = 0.035m, k = 250 N/m

= √k/m = √500 = 22.36 rad.s-1

ET = ½mω2xo2 = ½(0.5)(22.36)2(0.035)2 = 0.153J.

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• A mass-spring system oscillates with an amplitude of 3.5cm.  If the force constant of the spring of 250Nm-1 and the mass is 0.5kg, determine(a) the total energy of the system,(b) the maximum speed of the mass(c) the maximum acceleration

(b) Maximum speed when x = 0 (Ek = max)

At this point, the total energy is all kinetic

Ek = 0.153 J = ½mv2

v2 = (2 x 0.153)/0.5 = 0.612v = 0.78m.s-1

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• A mass-spring system oscillates with an amplitude of 3.5cm.  If the force constant of the spring of 250Nm-1 and the mass is 0.5kg, determine(a) the total energy of the system,(b) the maximum speed of the mass(c) the maximum acceleration

(c) Maximum acceleration (force) when x = xo

F = -kx = -250 x 0.035 = 8.75 N

a = F/m = 8.75/0.5 = 17.5 m.s-2

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Last lesson

• Considering energy changes during SHM• Applying Ek = ½mω2(xo

2 – x2)

ET = ½mω2xo2

Ep = ½mω2x2

where ω = 2πf = 2π/T

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This lesson

• Know what is meant by damping• Describe damped oscillations• Know what is meant by natural frequency

and forced oscillation• Describe graphically forced oscillations

near the natural frequency• Know what is meant by resonance and

describe examples

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Damping

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Damping

In most real oscillating systems energy is lost through friction.

The amplitude of oscillations gradually decreases until they reach zero. This is called damping

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Heavy Damping

Sometimes the damping is so large that the motion no longer resembles SHM

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Overdamped

The system takes a long time to reach equilibrium

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Underdamped

The system makes several oscillations before coming to rest

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Critical damping

Equilibrium is reached in the quickest time

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Forced vibrations - resonance

YouTube - Tacoma Bridge Disaster

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Natural frequency

All objects have a natural frequency that they prefer to vibrate at.

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Forced vibrations

If a force is applied at a different frequency to the natural frequency we get forced vibrations

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Resonance

If the frequency of the external force is equal to the natural frequency we get resonance

YouTube - Ground Resonance - Side View

YouTube - breaking a wine glass using resonance

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Let’s try some more questions!