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SHM -1
Springs ACT
• Hooke’s Law: The force exerted by a spring is
proportional to the distance the spring is stretched
or compressed from its relaxed position.
– FX = -k x
Where x is the displacement from
the relaxed position and k is the
constant of proportionality.
What is force of spring when it is stretched
as shown below.
A) F > 0
B) F =relaxed
0 position C) F < 0
FX = - kx < 0
x
x>0
x=0
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Spring ACT
A mass on a spring oscillates back & forth with simple
harmonic motion of amplitude A. A plot of
displacement (x) versus time (t) is shown below. At
what points during its oscillation is the magnitude of
the acceleration of the block biggest?
1. When x = +A or -A (i.e. maximum displacement) CORREC
T
2. When x = 0 (i.e. zero displacement)
3. The acceleration of the mass is constant
F=ma
x
+A
t
-A
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Potential Energy in Spring
• Force of spring is Conservative
– F = -k x
– W = -1/2 k x2
Force
work
x
– Work done only depends on initial and
final position
– Define Potential Energy Uspring = ½ k
x2
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Oscillations
Oscillations (whether sinusoidal or otherwise) have some
common characteristics:
1. They take place around an equilibrium position;
2. The motion is periodic and repeats with each cycle.
Periodic Motion
Period: time required for one cycle of periodic motion
Frequency: number of oscillations per unit time
The frequency unit is called
a hertz (Hz):
Frequency and Period
f  1/ T
and
T  1/ f
f is the frequency T is the period (units: s)
(units: Hz  oscillations per second)
Example: Radio Station
Frequency and Period
What is the oscillation period of an FM radio station
that broadcasts at 100 MHz?
f  100 MHz  1.0 108 Hz
1
8
T  1/ f 

1.0

10
s  10 ns
8
1.0 10 Hz
Note that 1/Hz = s
Simple Harmonic Motion
A spring exerts a restoring
force that is proportional to
the displacement from
equilibrium:
Simple Harmonic Motion
A mass on a spring has a displacement as a function of time that is a sine
or cosine curve:
A is called the amplitude
of the motion.
Simple Harmonic Motion
If we call the period of the motion T (this is the time to complete one full
cycle) we can write the position as a function of time as:
It is then straightforward to show that the position at time t + T is the
same as the position at time t (one period earlier), as we would expect.
Connections between Uniform Circular
Motion and Simple Harmonic Motion
An object in simple harmonic
motion has the same motion as one
component of an object in uniform
circular motion:
Connections between Uniform Circular
Motion and Simple Harmonic Motion
Uniform circular
motion projected into one
dimension is simple
harmonic motion (SHM).
Consider a particle
rotating ccw, with the
angle f increasing
linearly with time:
x  A cos f

f
, so f  t if f  0 at t  0.
t
x(t )  A cos t
Connections between Uniform Circular
Motion and Simple Harmonic Motion
Here, the object in circular motion has an angular speed of
where T is the period of motion of the object in simple harmonic motion.
Connections between Uniform Circular
Motion and Simple Harmonic Motion
The position as a function of time:
The angular frequency:
Connections between Uniform Circular
Motion and Simple Harmonic Motion
The velocity as a function of time:
And the acceleration:
Both of these are found by taking components of the circular motion
quantities.
The Period of a Mass on a Spring
Since the force on a mass on a spring is proportional to the displacement, and
also to the acceleration, we find that
.
Substituting the time dependencies of a and x gives:
The Period of a Mass on a Spring
Therefore, the period is:
Mass+Spring
Simple Harmonic Motion
F  kx  max
k
ax   x
m
In simple harmonic motion (SHM), the acceleration, and thus
the net force, are both proportional to and oppositely directed from
the displacement from the equilibrium position.
1 
frequency  f  
T 2
x(t )  A cos t   
A = amplitude
 = angular frequency
 = phase
SHM Prototype Experiment
Consider Fig. (a). An airtrack glider attached to a
spring. The glider is pulled a
distance A from its rest
position and released.
Fig. (b) shows a graph of
the motion of the glider, as
measured each 1/20 of a
second.
The graphs on the right
show the position and velocity
of the glider from the same
measurements. We see that
A=0.17 m and T=1.60 s.
Therefore the oscillation
frequency of the system is f =
0.625 Hz
Two Oscillating Systems
The diagram shows two identical
masses attached to two identical
springs and resting on a horizontal
frictionless surface. Spring 1 is
stretched to 5 cm, spring 2 is
stretched to 10 cm, and the masses
are released at the same time.
Which mass reaches the
equilibrium position first?
Because k and m are the same, the
systems have the same period, so
they must return to equilibrium at the
same time.
The frequency and period of SHM
are independent of amplitude.
Clicker Question 1
Shown are two mass
+ spring systems. The
blocks have the same
mass.
When set into
oscillation, what is the
relation between the
oscillation periods T1,2 of
the two systems?
(a) T1>T2 (b) T1=T2 (c) T1<T2
(d) Need to know m and k to answer
Example: A Block on a Spring
A 2.00 kg block is attached to a spring as shown.
The force constant of the spring is k = 196 N/m.
The block is held a distance of 5.00 cm from
equilibrium and released at t = 0.
(a) Find the angular frequency , the frequency f, and the period T.
(b) Write an equation for x vs. time.
 (9.90 rad/s)

 1.58 Hz
2
2
k
(196 N/m)


 9.90 rad/s
m
(2.00 kg)
f 
T  1/ f  0.635 s
A  5.00 cm and   0
x  (5.00 cm)cos (9.90 rad/s)t 
Example: A System in SHM
An air-track glider is attached to a spring,
pulled 20 cm to the right, and released
at t-=0. It makes 15 complete
oscillations in 10 s.
a. What is the period of oscillation?
b. What is the object’s maximum speed?
c. What is its position and velocity at t=0.80 s?
15 oscillations
f 
10 s
 1.5 oscillations/s  1.5 Hz
x  A cos
T  1/ f  0.667 s
2 A 2 (0.20 m)
vmax 

 1.88 m/s
T
(0.667 s)
2 t
2 (0.80 s)
 (0.20 m) cos
 0.062 m  6.2 cm
T
(0.667 s)
v  vmax sin
2 t
2 (0.80 s)
 (1.88 m/s)sin
 1.79 m/s  179 cm/s
T
(0.667 s)
Example: Finding the Time
A mass, oscillating in simple harmonic motion,
starts at x = A and has period T.
At what time, as a fraction of T, does the mass
first pass through x = ½A?
2 t
x  A  A cos
T
1
2
T
T  1
1  1 
t
cos  2  
 6T
2
2 3
The Phase Constant
But what if f is not zero at t=0?
f  t  f0
x(t )  A cos t  f0 
v(t )   A sin t  f0 
 vmax sin t  f0 
Set t  0:
x0  x(0)  A cos f0
v0 x  v(0)   A sin f0
A phase constant f0 means that the
rotation starts at a different point on the
circle, implying different initial conditions.
SHM Initial Conditions

k
m
x(t )  A cos t   
x0  x(t  0)  A cos 
v0 x  vx (t  0)   A sin 
v0 x
   arctan
 x0
a0 x  ax (t  0)   2 A cos    2 x0
x(t )  x(t  T )  x(t  nT )
Since  
k
,
m
f 
1
2
T  2 or   2 / T
k
m
and T  2
m
k
t


x  A cos  2   
 T

***Energy ***
• A mass is attached to a spring and set to
motion. The maximum displacement is x=A
 SWnc = K + U
–
0 = K + U or Energy U+K is constant!
Energy = ½ k x2 + ½ m v2
– At maximum displacement x=A, v = 0
PE
S
Energy = ½ k A2 + 0
– At zero displacement x = 0
Energy = 0 + ½ mvm2
Since Total Energy is same
0
½ k A2 = ½ m vm2
m
vm = sqrt(k/m) A
x=0
x
x
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Preflight 3+4
A mass on a spring oscillates back & forth with simple
harmonic motion of amplitude A. A plot of
displacement (x) versus time (t) is shown below. At
what points during its oscillation is the total energy
(K+U) of the mass and spring a maximum? (Ignore
gravity).
1. When x = +A or -A (i.e. maximum displacement)
2. When x = 0 (i.e. zero displacement)
CORRECT
3. The energy of the system is constant.
x
+A
t
-A
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Preflight 1+2
A mass on a spring oscillates back & forth with simple
harmonic motion of amplitude A. A plot of
displacement (x) versus time (t) is shown below. At
what points during its oscillation is the speed of the
block biggest?
1. When x = +A or -A (i.e. maximum displacement)
CORREC
2. When x = 0 (i.e. zero displacement)
T
3. The speed of the mass is constant
“There is no potential energy at x=0 since U=1/2kx^2=0,
therefore allowing all the energy of the spring to be
allocated toward KE .
x
+A
t
-A
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Simple Harmonic Motion:
x(t) = [A]cos(t)
v(t) = -[A]sin(t)
a(t) = -[A2]cos(t)
xmax = A
vmax = A
amax = A2
OR
x(t) = [A]sin(t)
v(t) = [A]cos(t)
a(t) = -[A2]sin(t)
Period = T (seconds per cycle)
Frequency = f = 1/T (cycles per second)
Angular frequency =  = 2f = 2/T
For spring: 2 = k/m
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Example
A 3 kg mass is attached to a spring (k=24 N/m).
It is stretched 5 cm. At time t=0 it is released
and oscillates.
Which equation describes the position as a
function of time x(t) =
A) 5 sin(t)
B) 5 cos(t)
C) 24sin(t)
D) 24 cos(t) E) -24 cos(t)
We are told at t=0, x = +5 cm. x(t) = 5 cos(t) only
one that works.
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Example
A 3 kg mass is attached to a spring (k=24 N/m).
It is stretched 5 cm. At time t=0 it is released
and oscillates.
What is the total energy of the block spring
system?
A) 0.03 J
B) .05 J
C) .08 J
E=U+K
At t=0, x = 5 cm and v=0:
E = ½ k x2 + 0
= ½ (24 N/m) (5 cm)2
= 0.03 J
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Example
A 3 kg mass is attached to a spring (k=24 N/m).
It is stretched 5 cm. At time t=0 it is released
and oscillates.
What is the maximum speed of the block?
A) .45 m/s
B) .23 m/s
C) .14 m/s
E=U+K
When x = 0, maximum speed:
E = ½ m v2 + 0
.03 = ½ 3 kg v2
v = .14 m/s
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Example
A 3 kg mass is attached to a spring (k=24 N/m).
It is stretched 5 cm. At time t=0 it is released
and oscillates.
How long does it take for the block to return to
x=+5cm?
A) 1.4 s
B) 2.2 s
C) 3.5 s
 = sqrt(k/m)
= sqrt(24/3)
= 2.83 radians/sec
Returns to original position after 2  radians
T = 2  /  = 6.28 / 2.83 = 2.2 seconds
• Springs
Summary
– F = -kx
– U = ½ k x2
  = sqrt(k/m)
• Simple Harmonic Motion
– Occurs when have linear restoring force
F= -kx
– x(t) = [A] cos(t)
or
[A] sin(t)
– v(t) = -[A] sin(t) or
[A] cos(t)
– a(t) = -[A2] cos(t) or
-[A2] sin(t)