Simple Harmonic Motion(SHM)
• Vibration (oscillation)
• Equilibrium position – position of the natural
length of a spring
• Amplitude – maximum displacement
Period and Frequency
• Period (T) – Time for one complete cycle (back
to starting point)
• Frequency (Hz) – Cycles per second
• Angular Velocity/Frequency (rad/s), w – FIND
THIS FIRST
f=1
T
T=1
f
w = 2pf
Period and Frequency
A radio station has a frequency of 103.1 M Hz.
What is the period of the wave?
103.1 M Hz
1X106 Hz = 1.031 X 108 Hz
1M Hz
T = 1/f = 1/(1.031 X 108 Hz) = 9.700 X 10-9 s
Cosines and Sines
• Imagine placing a
pen on a vibrating
mass
• Draws a cosine
wave
1
Starting at the Amplitude
• Used if spring is pulled out (or compressed) to
full position
x(t) = A cos2pt = A cos2pft = A coswt
T
v(t) = -Awsinwt
vmax = Aw
A = Amplitude
t = time
T = period
f = frequency
A mass starts at x=A. Using only variables,
calculate at what time as a fraction of T that the
object passes through ½ A.
x(t) = A cos2pt
T
An air-track glider is attached to a spring. It is
pulled 20.0 cm to the right and makes 15
oscillations in 10.0s
a. Calculate the period (0.667 s)
b. Calculate the angular velocity (w) (9.42 rad/s)
c. Calculate the maximum speed (+1.88 m/s)
d. Calculate the speed and position at t = 0.800s
(6.25 cm, -1.79 m/s)
x = A coswt
Velocity is the
derivative of
position
v = -Awsinwt
Acceleration is
the derivative of
velocity
a = -Aw2coswt
A loudspeaker vibrates at 262 Hz (middle C). The
amplitude of the cone of the speaker is 1.5 X 10-4
m.
a. Write the equation to describe the position of the
cone over time. (x = (1.5 X 10-4 m) cos(1650
rad/s)t)
b. Calculate the position at t = 1.00 ms. (-1.2 X 10-5
m)
c. Calculate the maximum velocity and
acceleration (0.25 m/s, 410 m/s2 )
Find the amplitude, frequency and period of motion
for an object vibrating at the end of a spring that
follows the equation:
x = (0.25 m)cos p t
8.0
(0.25 m, 1/16 Hz, 16 s)
2
The Phase Constant
Find the position of the object after 2.0 seconds.
x = (0.25 m)cos p t
8.0
x = (0.25 m)cos p
4.0
• Do not always start at the Amplitude
• Can start your observations at any time
o is the starting angle using the circle model
x(t) = A cos(wt + o)
v(t) = -A w sin(wt + o)
x = 0.18 m
An object on a spring oscillates with a period of
0.800 s and an amplitude of 10.0 cm. At t=0, it
is 5.0 cm to the left of equilibrium and moving
to the left.
a. Calculate the phase constant (in radians) from
the initial conditions. (2/3p rad)
b. Calculate the position at t = 2.0 s (5.0 cm)
c. Calculate the velocity at t = 2.0 s (68 cm/s)
d. What direction is the object moving at 2 s?
(right)
An air-track glider attached to a spring oscillates
with a period of 1.50s . At t=0 s the glider is
5.10 cm left of the equilibrium position and
moving to the right at 37.0 cm/s.
a. Calculate the phase constant. (+1.05 rad)
b. Calculate the amplitude (-10.3 cm)
c. Calculate the position and velocity at t=5.00 s.
(+10.3, -0.12 cm/s)
The initial position and velocity of a block
moving in SHM with period T=0.25 s are x(0)
= 5.0 cm and v(0) = 218 cm/s.
a. Calculate the angular velocity w. (25.1 rad/s)
b. Calculate the amplitude (10.0 cm)
c. Calculate the phase constant (in radians) from
the initial conditions. (-1.05 rad)
Forces on a Spring
• Extreme Position (Amplitude)
– Force at maximum
– Velocity = 0
• Equilibrium position
– Force = 0
– Velocity at maximum
3
d2x + kx = 0
dt2
m
The Equation of Motion
F = ma
F = -kx
ma = -kx
a = -kx
m
2
d x = -kx
dt2 m
also
a = dv = d2x
dt
dt2
x(t)
dx/dt
d2x/dt2
Equation of Motion
= A cos(wt + j)
= -wA sin(wt + j)
= -w2A cos(wt + j)
-w2A cos(wt + j) + k A cos(wt + j) = 0
m
Energy and Springs
All PE
• KE = ½ mv2
• PE = ½ kx2
• Maximum PE = ½ kA2
All KE
Law of conservation of Energy
½ kA2 = ½ mv2+ ½ kx2
All PE
Also
w = (k/m)½
A 0.50 kg mass is connected to a light spring
with a spring constant of 20 N/m.
a. Calculate the total energy if the amplitude is 3.0
cm. (9 X 10-3 J)
b. Calculate the maximum speed of the mass (0.19
m/s)
c. Calculate the potential energy and kinetic energy
at x = 2.0 cm (U = 4 X 10-3 J, K = 5 X 10-3 J)
d. At what position is the speed 0.10 m/s? (+ 2.6
cm)
Some KE and Some PE
A mass of 4.00 kg is attached to a horizontal
spring with k = 100 N/m. It is displaced 10.0
cm from equilibrium and released.
a. Calculate the period. (1.25 s)
b. Calculate the angular velocity w. (5.00 rad/s)
c. Calculate the maximum velocity (0.500 m/s)
d. Calculate the velocity when x = 5.0 cm (0.433
m/s)
4
A spring stretches 0.150 m when a 0.300 kg mass
is suspended from it.
a. Find the spring constant. (19.6 N/m)
b. The spring is now stretched an additional 0.100
m and allowed to oscillate. What is the
maximum velocity? (0.808 m/s)
c. Calculate the velocity at x = 0.0500 m (0.700
m/s)
d. What is the maximum acceleration? (6.53 m/s2)
Trigonometry and SHM
• Ball rotates on a table
• Looks like a spring from the side
• Amplitude = Radius
T = 2p m
k
A 500 g block is pulled 20 cm on a spring and
released. It has a period of 0.800 s . At what
positions is the block’s speed 1.0 m/s? (Hint:
use w = \/k/m )
• Period depends only on mass and spring constant
• Amplitude does not affect period
vmax = 2pAf
or
vmax = 2pA
T
w = \/k/m
w = \/k/m
f= 1
T
What is the period and frequency of a 1400 kg car
whose shocks have a k of 6.5 X 104 N/m after it
hits a bump?
w=
k =
m
An insect (m=0.30 g) is caught in a spiderweb
that vibrates at 15 Hz.
a. What is the spring constant of the web? (2.7
N/m)
b. What would be the frequency for a lighter insect,
0.10 g? Would it be higher or lower? (26 Hz)
6.81 rad/s
w = 2pf
ANS: 1.09 Hz
5
Vertical Motion of a Spring
• Gravity is ALWAYS acting on the spring and
mass consistently
• Only need to use it to calculate the spring
constant
• F = -ky (using y for vertical rather than x)
An 83 kg student hangs from a bungee cord with
spring constant 270 N/m. He is pulled down to
a position 5.0 m below the unstretched length
of the bungee, then released.
a. Calculate the equilibrium length of the
bungee/student (3 m)
b. Calculate the Amplitude (2 m)
c. Calculate the position and velocity 2.0 s later.
(1.8 m, -1.6 m/s)
At t = 0, a 5000 g block is moving to the right
and is at 15 cm. Its maximum displacement is
25 cm at 0.30 s.
a. Calculate the phase constant
b. Calculate the angular velocity
c. Calculate the time and velocity when the mass is
at x = 20 cm
d. Sketch a graph of the motion, including the
phase constant and period
The Pendulum
• Pendulums follow SHM only
for small angles (<15o)
• The restoring force is at a
maximum at the top of the
swing.
q
Fr = restoring Force
Remember the circle (360o = 2p rad)
q
L
q=x
L
Fr = -mgq
ma = -mgq
a = -gq
d2s = -gq
Dt2
Fr = -mgsinq
at small angles sinq = q
Fr = -mgq
q =s
L
x
q
L
mg
The Pendulum and the Equation of motion
(small angle approximation)
x
6
d2s = -gs
dt2
L
(q = s/L)
s = Acos(wt + )
d2s = -w2Acos(wt + )
dt2
T = 2p L
g
The Period and Frequency of a pendulum depends
only on its length
-w2Acos(wt + ) = -g Acos(wt + )
L
Swings and the Pendulum
• To go fast, you need a high frequency
• Short length (tucking and extending your legs)
f= 1 g
2p L
Consider a grandfather clock with a 1.0 m long
pendulum
a. Calculate the period of? (2.0 s)
b. Estimate the length of the pendulum of a
grandfather clock that ticks once per second (T =
1.0 s). (0.25 m)
decrease the denominator
A 300 g mass on a 30 cm long string swings at a
speed of 0.25 m/s at its lowest point.
a. Calculate the period
b. Calculate the angular velocity
c. Calculate the maximum angle that the pendulum
reaches (HINT: Use the triangle and the small
angle approximation).
Physical Pendulum
• Center of mass is in the middle (sin q)
t = -Mglsinq
t = -Mglq
(small angle approximation)
7
t = Ia
a = d2q
dt2
t = I d 2q
dt2
t = -Mglq
d2q + Mglq = 0
dt2
I
d2x + k x = 0
dt2 m
I d2q = -Mglq
dt2
2
d q + Mglq = 0
dt2
I
A student swings his leg which is 0.90 m long.
Assume the center of mass is halfway down the
leg.
a. Write the equation for the moment of inertia of
the leg (I = ML2/3)
b. Substitute this into the period formula to
calculate the period (1.6 s)
c. Calculate the frequency (0.64 Hz)
A Christmas ball has a radius R and a moment
of inertia of 5/3MR2 when hung by a hook.
The ball is given a slight tap and rocks back
and forth.
a. Derive the formula for the period of oscillation.
Assume the center of mass is at the center of the
ball.
b. Insert a reasonable radius into your equation to
estimate the period.
A nonuniform 1.0 kg physical pendulum has a
center of mass 42 cm from the pivot. It
oscillates with a period of 1.6 s.
a. Calculate the moment of inertia (0.27 kg m2)
Damped Harmonic Motion
•Most SHM systems slowly stop
•For car shocks, a fluid “dampens” the motion
8
Resonance: Forced Vibrations
• Can manually move a spring (sitting on a car and
bouncing it)
• Natural or Resonant frequency (fo)
• When the driving frequency f = fo, maximum
amplitude results
– Tacoma Narrows Bridge
– 1989 freeway collapse
– Shattering a glass by singing
Wave Medium
• Mechanical Waves
– Require a medium
– Water waves
– Sound waves
– Medium moves up and down but wave moves
sideways
• Electromagnetic Waves
– Do not require a medium
– EM waves can travel through the vacuum of space
Parts of a wave
•
•
•
•
•
•
Crest
Trough
Amplitude
Wavelength
Frequency (cycles/s or Hertz (Hz))
Velocity
v = lf
9
•How many complete waves are shown above?
•What is the wavelength of light shown above?
2. a) 3.3 s b) 0.30 Hz
c) 0.25 m d) 0.48 m/s
10. a)
b) -2/3p rad, 0 rad, 2/3p rad,
4/3p rad
12. 5.48 N/m
14.a) 2.0 cm
b) 0.63 s c) 5.0 N/m d) -1/4p rad
e) 1.41 cm, 14.1 cm/s f) 20 cm/s g) 0.001J
h) 1.46 cm/s
16. a) 24.5 N/m
b) 0.90 s c) 0.70 m/s
18. a) 0.169 kg
b) 0.57 m/s
20.a) 0.10 rad b) 0.796 Hz
c) p rad
d) 0.392 m e) -0.10 rad
f) 0.084 rad
22. a) 2.0 s
32. a) 115 cm
34. 0.41 s
40.a) 6.4 cm
d) 0.283 m/s
42. 1.02 m/s
44. a) 3.2 Hz
58. 0.66 s
b) 8.86 m/s2
b) 5/6p rad
c) -100 cm
b) 160 cm/s2
c) -6.4 cm
b) 0.0707 m, 7.07 cm
c) 5.0 J
55)w = (g/L)1/2 = (9.8/1)1/2 = 3.13 rad/s
x = Acos (wt + j)
-4o = 8ocos (3.13t + 0)
8o
cos (3.13t) = -0.5
-4o
t = 0.67 s
10