Download Simple Harmonic Motion

Simple Harmonic Motion
A pendulum swinging from side to side is an
example of both periodic and simple harmonic
motion.
Periodic motion is when an object continually
moves back and forth over a definite path in
equal intervals of time.
Simple harmonic motion (SHM) is when the
acceleration of an object is proportional to the
displacement of the object from its equilibrium
position and is always directed toward the
equilibrium position.
Free-body diagram of a pendulum in its
equilibrium position:
T
•
Fw
T is the tension in the string and Fw
is the weight of the pendulum bob.
ΣF = 0
T = Fw
The free-body diagram at the amplitude:
T
Fr
θ
•
θ
F1
Fw
where Fw is the weight of the pendulum bob, Fr
is the restoring force, T is the tension in the
string, and F1 is the other component of Fw.
In order for the restoring force (Fr) to be
proportional to the displacement, 0 < θ <≈ 10°,
because θ ≈ sin θ for angles less than 10°.
The Vocabulary of a Pendulum
Amplitude is the maximum displacement from
the equilibrium position (m).
Frequency is the number of complete vibrations
usually given in /s or hz (1 hz = 1 cycle/s).
Period is the time needed for one complete
vibration (s).
Characteristics of a Pendulum
In the absence of air resistance, the period of a
pendulum is independent of the mass of the
pendulum bob.
The period of a pendulum is independent of the
amplitude if 0 < θ <≈ 10°.
The period of a pendulum is directly
proportional to the square root of its length.
The period of a pendulum is inversely
proportional to the square root of the
acceleration due to gravity.
The period of a pendulum is given by
T = 2π(l/g)1/2
where T is the period usually in s, l is the length
usually in m, and g = 9.80 m/s2 (dependent on
location).
Pendulum Problems
On the top of a small hill a pendulum 1.45 m
long has a period of 2.47 s. What is g for this
location?
T = 2.47 s
l = 1.45 m
T = 2π(l/g)1/2
g = 4π2l/T2 = 4 × 3.142 × 1.45 m/(2.47 s)2
g = 9.38 m/s2
Another Pendulum Problem
A 0.70 kg pendulum bob at the end of a 0.85 m
string is pulled back 0.90 m and released.
m = 0.70 kg
l = 0.85 m
Δx = 0.10 m
(a) What is the acceleration of the pendulum
bob at the instant it is released?
The free-body diagram at the amplitude:
T
Fr
θ
•
θ
F1
Fw
θ ≈ sin θ for small angles, therefore
θ = sin-1(0.10/0.80) and sin θ = Fr/Fw.
Fr = Fw × sinθ = mgsinθ
Fr = 0.70 kg × 9.80 m/s2 × 0.10 m/0.85 m
Fr = 0.81 N
a = Fr/m = 0.81 N/0.70 kg = 1.2 m/s2
(b) What is the acceleration as the pendulum
passes through the equilibrium position?
0 because Fr = 0.
(c) What is the net force at the instant of
release?
Fnet = Fr = 0.81 N (See part (a)).
(d) What is the period of the pendulum?
T = 2π(l/g)1/2
T = 2 × 3.14 × ((0.85 m/9.80 m/s2))1/2
T = 1.9 s
(e) What is the period of the pendulum if it was
in a freely falling elevator?
T = 2π(l/g)1/2
In free fall, the “effective g” equals zero so
the period would be infinite which correlates
to no swing.
Wrap Up Questions
What is the distance traveled by an object
moving with simple harmonic motion during a
time that is equal to its period?
The distance would be 2 × Δxmax where xmax is
the amplitude of vibration.
A pendulum is hung from the ceiling of a
stationary elevator. After determining the
period of the pendulum, the elevator
accelerates upward, accelerates downward,
and finally moves downward with a constant
velocity. The period is determined while
undergoing these motions.
Did the period of the pendulum change while
the elevator accelerated upward?
During the upward acceleration, the apparent
value for g increases. Remember, during the
upward acceleration, there has to be a net force
acting upward in accordance with Newton’s 2nd
Law.
The period of a pendulum is given by:
T = 2 × π × (l/g)1/2.
An increase in g will decreases the period, T.
Did the period of the pendulum change while
the elevator accelerated downward?
During the downward acceleration, the apparent
value for g decreases.
The period of a pendulum is given by:
T = 2 × π × (l/g)1/2.
A decrease in g will increases the period, T.
Did the period of the pendulum change while
the elevator moved downward with a constant
velocity?
While the elevator moves downward with a
constant velocity, the value of g remains the
same.
The period of a pendulum will remain the same.
In the case of free fall, the pendulum does not
oscillate.