Measuring Harmonic
Motion
Amplitude
• Maximum displacement from the equilibrium
position
Period
• Period is represented by T
• T= is the time it takes for a complete
cycle of motion measured in seconds
• During T , the displacement is zero
Frequency
• Frequency is represented by f
• f= is the number of complete cycles
through a unit of time
• SI units of s-1 (Hertz)
Frequency and Period
• Frequency and period are inversely
proportional
F= 1/T
and T= 1/F
F is # of cycles/per unit time
T is time /cycle
Period of a Simple Pendulum
• Depends on string length and free fall
acceleration
• Ex. #1 - 2 pendulums with different
masses, have the same period
• Ex. #2 - 2 pendulums have different
amplitudes, have the same period
Period of a Simple Pendulum
• T = 2 ∏ √ (L/g)
• Period = 2 ∏ X √ (length divided by 9.81)
Period of a Simple Pendulum
• Why does the length of the string
affect the period?
• The shorter the pendulum, the
smaller the arc to travel through
Period of a Simple Pendulum
• Why doesn’t the mass and amplitude
affect the period?
– Heavier mass has a greater restoring
force and larger force to achieve the same
acceleration
– Free fall is the same, regardless of the
mass
Example 1
You need to know the height of a tower, but
darkness obscures the ceiling. You note that a
pendulum extending from the ceiling almost
touches the floor and that its period is 12 seconds.
How tall is the tower?
L = 36 m
Example 2
If the length of the string in the previous problem
was halved, what would the period be?
8.5 seconds
Example 3
The cables of a trapeze are approximately 3.5 m
long. What is the period of an acrobat swinging on
this trapeze?
3.8 seconds
Example 4
What is the frequency of an acrobat swinging on the
trapeze described in Example 2?
0.26 Hz
Example 5
Calculate the period and frequency of a 3.500m long
pendulum at the following locations:
The North Pole, where g= 9.832 m/s2
3.749 s; 0.2667 Hz
Example 6
Calculate the period and frequency of a 3.500m long
pendulum at the following locations:
The Chicago, where g= 9.803 m/s2
3.754 s; 0.2664 Hz
Example 7
Calculate the period and frequency of a 3.500m long
pendulum at the following locations:
The Jakarta, Indonesia, where g= 9.782 m/s2
3.758 s; 0.2661 Hz
Period of a Mass-Spring System
• Restoring force is determined by the
displacement
• Magnitude of the mass, does not affect
the restoring force
• Therefore, mass increased both force
on the bob and the bob’s inertia
Period of a Mass-Spring System
• Heavy masses have smaller
accelerations, than light masses
• Heavy masses will take more time to
complete one cycle of motion
• Heavy masses = greater period
Period of a Mass-Spring System
• Greater the spring constant = greater
force and acceleration
• Period will decrease
Period of a Mass-Spring System
• Greater the spring constant = greater
force and acceleration
• Period will decrease
Period of a Mass-Spring System
• T = 2 ∏ √ (m/k)
• Period = 2 ∏ √ (mass divided by spring
constant)
Example 8
The body of a 1250 kg car is supported on a frame by
four springs, each of which has a spring constant
of 20,000 N/m. Two people riding in the car have
combined mass of 153 kg. Find the period of
vibration of the car when it is driven over a
pothole in the road.
T = 0.84s
Example 9
What is the frequency of the vibration from Example
8?
1.2 Hz
Example 10
If two more people enter the car, the occupants total
mass increases to 255 kg. What is the period of
vibration of the car now?
0.87 s
Example 11
How long will it take the car in Example 10 to
complete three vibrations?
2.6 s
Example 12
A pendulum swings from maximum displacement
on one side of equilibrium to maximum
displacement on the opposite side of the
equilibrium. If the pendulum swings through a
total of 24 °, what is the amplitude of this
vibration
12 °