Download Simple Harmonic Motion

10.2 Simple Harmonic Motion and the Reference Circle
DISPLACEMENT
10.2 Simple Harmonic Motion and the Reference Circle
A – radius of the circle
𝒙 – displacement of ball from
equilibrium position.
x  A cos  A cos t
10.2 Simple Harmonic Motion and the Reference Circle
amplitude A: the maximum displacement
period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
10.2 Simple Harmonic Motion and the Reference Circle
DISPLACEMENT
x  A cos   A cos t  A cos(2f )t
Velocity

Using the reference circle model
to determine the velocity of an
object:

𝑣𝑥 – 𝑥 component of 𝑣𝑡 = −𝑣𝑡𝑠𝑖𝑛𝜃


Negative sign indicates direction of
𝑣𝑥
Where:
 𝜃 = 𝜔𝑡
 𝑣𝑡 = 𝑟𝜔 (since r = A) then: 𝑣𝑡 = 𝐴𝜔

Therefore the velocity in SHM is
given by:
v x  vt sin    A sin t
(ω in rad/s)
Oscillating Spring/mass Systems
Maximum
velocity is
experienced at
the equilibrium
point, where
Felastic = 0.
 At maximum
displacement,
the velocity is
zero.

© 2002 HRW
10.2 Simple Harmonic Motion and the Reference Circle
VELOCITY
v x  vT sin    
A sin t
vmax
Acceleration

Using the reference circle model
to determine the acceleration of
an object:

The centripetal acceleration of the
ball points towards the center, 𝑎𝑐

𝑎𝑥 – 𝑥 component of 𝑎𝑐 = −𝑎𝑐𝑐𝑜𝑠𝜃


Negative sign indicates direction of
𝑎𝑥
Where:
 𝑎𝑐 = 𝑟𝜔2 (since r = A) then: 𝑎𝑐 = 𝐴𝜔2

Therefore the acceleration in
SHM is given by:
a x   ac cos    A 2 cos t
(ω in rad/s)
10.2 Simple Harmonic Motion and the Reference Circle
ACCELERATION
ax  ac cos   
A 2 cos t
amax
Consider the graph shown for the position of a ball attached to a
spring as it oscillates in simple harmonic motion. At which of the
following times is the ball at its equilibrium position?
a) 0 s only
b) 2 s only
c) 4 s only
d) at 0 s and 8 s
e) at 0 s, 4 s, and 8 s
Which one of the following statements concerning the total
mechanical energy of a harmonic oscillator at a particular point in
its motion is true?
a) The mechanical energy depends on the acceleration at that point.
b) The mechanical energy depends on the velocity at that point.
c) The mechanical energy depends on the position of that point.
d) The mechanical energy does not vary during the motion.
e) The mechanical energy is equal to zero joules if the point is the
equilibrium point.
10.3 Energy and Simple Harmonic Motion

A stretched or compressed spring has elastic
potential energy and therefore it can do work.
10.3 Energy and Simple Harmonic Motion

From the work equation:

𝑊𝑒𝑙𝑎𝑠𝑡𝑖𝑐 = 𝐹𝑐𝑜𝑠𝜃 𝑑

From the diagram below: 𝑑 = 𝑥𝑜 − 𝑥𝑓

The magnitude of the spring force given by 𝐹𝑥 = −𝑘𝑥 is
not constant and changes from 𝑘𝑥𝑜 to 𝑘𝑥𝑓 therefore the
magnitude of the average force is:

1
𝐹𝑥 = 2 (𝑘𝑥𝑜 + 𝑘𝑥𝑓 )
Welastic  F cos s  12 kxo  kxf cos 0 xo  x f 
Welastic 
1
2
kxo2  12 kx 2f
10.3 Energy and Simple Harmonic Motion

The elastic potential energy is the energy that a
spring has by virtue of being stretched or
compressed. For an ideal spring, the elastic
potential energy is
PEelastic  12 kx2

SI unit: Joule (J)
10.3 Energy and Simple Harmonic Motion
Example 5: Adding a Mass to a Simple
Harmonic Oscillator

A 0.20-kg ball is attached to a vertical spring. The spring
constant is 28 N/m. When released from rest, how far does
the ball fall before being brought to a momentary stop by
the spring?

G: m=0.20 kg, k=28N/m, hf=0m, vo=0m/s,vf=0m/s
U: ho=?
E: ME f  MEo
2
2
2
2
2
2
S: 12 mv f  12 I f  mghf  12 ky f  12 mvo  12 Io  mgho  12 kyo




Spring extension: yf = ho
1
2
kho2  mgho
ho 
2mg
k
20.20kg 9.81m / s2 

 0.14 m
28 N/m