Simple Harmonic Motion 3 - PowerPoint PPT Presentation

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Simple Harmonic Motion 3. Learning Objectives. Book Reference : Pages 38-39. Simple Harmonic Motion 3. To investigate the relationship between circular motion and SHM To be able to calculate the displacement at any particular moment under different starting conditions

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Simple Harmonic Motion 3

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Simple Harmonic Motion 3

Learning Objectives

Book Reference : Pages 38-39

Simple Harmonic Motion 3

To investigate the relationship between circular motion and SHM

To be able to calculate the displacement at any particular moment under different starting conditions

To be able to calculate the velocity at any particular moment.

y

• Consider an object in circular motion... At any given time the x and y coordinates are given by:

• x = r cos

• y = r sin 

r cos 

r sin 

Circular Motion & SHM

r

x

Graphically the x coordinate changes as shown

x

+r

sinewavedisc.swf

-r

We have seen that the displacement of an object in SHM is described by a sinusoidal function. However, there are differences between where we start the motion at time t=0

Displacement 1

x

Maximum displacement A at time t=0 (cosine)

+A

-A

x

Zero displacement at time t=0 (sin)

+A

-A

Providing that we operate in radians,the displacement x of an object in SHM is given by:

x = A cos 2ft (for displacement A at time t=0)

x = A sin 2ft (for zero displacement at time t=0)

Where A is the maximum displacement

Calculator must be in radians for this to work

Displacement 2

• The velocity of an object in SHM is given by:

• v =  2f (A2 – x2)

• Where v is the velocity, f is the frequency, A is the maximum displacement and x is the displacement

• The velocity is considered positive if moving away from the equilibrium point and negative if moving towards it

• Note this will be a maximum when x=0

• vmax= 2fA

• Velocity

• In a similar way the maximum value for acceleration given by the acceleration equation we saw last lesson will be....

• Acceleration = - (2f)2 x displacement

• Accelerationmax = - (2f)2 A

• Where f is the frequency, and A is the maximum displacement .

Maximum Acceleration

• A spring oscillates in SHM with a period of 3s and an amplitude of 58mm. Calculate:

• The frequency [0.33 Hz]

• The maximum acceleration [0.25 ms-2]

• The displacement of an object oscillating in SHM changes with time and is described by

• X (mm) = 12 cos 10t where t is the time in seconds after the object’s displacement was at its maximum value. Determine:

• The amplitude [12mm]

• The time period [0.63s]

• The displacement at t = 0.1s [6.5mm]

Problems 1

• An object on a spring oscillating in SHM has a time period of 0.48s and a maximum acceleration of 9.8m s-2. Calculate:

• Its frequency [2.1 Hz]

• Its Amplitude [0.057m]

Problems 2

• An object oscillates in SHM with an amplitude of 12mm and a period of 0.27s. Calculate

• The frequency [3.7 Hz]

• Its displacement and direction of motion 0.1s, and 0.2s after the displacement was +12mm

• [-8.2mm towards maximum negative displacement]

• [-0.7mm towards maximum positive displacement]