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 = 1 w= 2pf
T f
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 1X106Hz = 1.031 X 108Hz 1M Hz
T = 1/f = 1/(1.031 X 108Hz) = 9.700 X 10-9 s
Cosines and Sines
• Imagine placing a pen on a vibrating mass
• Draws a cosine wave
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
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)
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
x = A coswt
v = -Awsinwt
a = -Aw2coswt
Velocity is the derivative of position
Acceleration is the derivative of velocity
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 10-4m) 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
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
x = 0.18 m
The Phase Constant
• Do not always start at the Amplitude
• Can start your observations at any time
ois the starting angle using the circle model
x(t) = A cos(wt + o) v(t) = -Aw sin(wt + o)
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. (2/3prad)
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)
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)
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)
Forces on a Spring
• Extreme Position (Amplitude) – Force at maximum
– Velocity = 0
• Equilibrium position – Force = 0
– Velocity at maximum
The Equation of Motion F = ma
F = -kx
ma = -kx
a = -kx also a = dv = d2x
m dt dt2
d2x = -kx dt2 m
d2x + kx = 0 Equation of Motion dt2 m
x(t) = A cos(wt + j) dx/dt = -wA sin(wt + j) d2x/dt2 = -w2A cos(wt + j)
-w2A cos(wt + j) + k A cos(wt + j) = 0 m
Energy and Springs
• KE = ½ mv2
• PE = ½ kx2
• Maximum PE = ½ kA2
Law of conservation of Energy
½ kA2= ½ mv2+ ½ kx2
Also w= (k/m)½
All PE
All KE
All PE
Some KE and Some PE
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 10-3J)
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 10-3J, K = 5 X 10-3J) d. At what position is the speed 0.10 m/s? (+ 2.6
cm)
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)
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)
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 )
Trigonometry and SHM
• Ball rotates on a table
• Looks like a spring from the side
• Amplitude = Radius
T = 2p m w= \/k/m k
f = 1 T
• Period depends only on mass and spring constant
• Amplitude does not affect period
vmax= 2pAf or vmax= 2pA T
w= \/k/m
What is the period and frequency of a 1400 kg car whose shocks have a k of 6.5 X 104N/m after it hits a bump?
w= k = 6.81 rad/s m
w= 2pf ANS: 1.09 Hz
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)
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)
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
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)
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= 2prad)
q= x L
Fr= -mgsinq
at small angles sinq = q
q L
x
q
The Pendulum and the Equation of motion
Fr= -mgq (small angle approximation) ma = -mgq
a = -gq d2s = -gq Dt2
q= s
d2s = -gs (q= s/L) dt2 L
s = Acos(wt + ) d2s = -w2Acos(wt + )
dt2
-w2Acos(wt + ) = -g Acos(wt + ) L
T = 2p L g
The Period and Frequency of a pendulum depends only on its length
Swings and the Pendulum
• To go fast, you need a high frequency
• Short length (tucking and extending your legs)
f = 1 g
2p L decrease the denominator
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)
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)
t = Ia a = d2q dt2 t = I d2q
dt2
t = -Mglq
I d2q = -Mglq dt2
d2q + Mglq= 0 dt2 I
d2q + Mglq= 0 dt2 I
d2x + k x = 0 dt2 m
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 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)
A Christmas ball has a radius R and a moment of inertia of 5/3MR2when 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.
Damped Harmonic Motion
•Most SHM systems slowly stop
•For car shocks, a fluid “dampens” the motion
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
•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) -2/3prad, 0 rad, 2/3prad,
4/3prad 12. 5.48 N/m
14.a) 2.0 cm b) 0.63 s c) 5.0 N/m d) -1/4prad 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) prad d) 0.392 m e) -0.10 rad f) 0.084 rad
22. a) 2.0 s b) 8.86 m/s2
32. a) 115 cm b) 5/6prad c) -100 cm 34. 0.41 s
40.a) 6.4 cm b) 160 cm/s2 c) -6.4 cm d) 0.283 m/s
42. 1.02 m/s
44. a) 3.2 Hz b) 0.0707 m, 7.07 cm c) 5.0 J 58. 0.66 s
55)w= (g/L)1/2 = (9.8/1)1/2= 3.13 rad/s x = Acos (wt + j)
-4o = 8ocos (3.13t + 0) cos (3.13t) = -0.5 t = 0.67 s
8o -4o
