Physics 1501: Lecture 25 Today ’ s Agenda. Homework #9 (due Friday Nov. 4) Midterm 2: Nov. 16 Topics Review of static equilibrium Oscillation Simple Harmonic Motion – masses on springs Energy of the SHO. Approach to Statics:. In general, we can use the two equations - PowerPoint PPT Presentation
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TopicsReview of static equilibriumOscillationSimple Harmonic Motion – masses on springs Energy of the SHO
Physics 1501: Lecture 25, Pg 2
Approach to Statics:Approach to Statics:
In general, we can use the two equations
to solve any statics problems.
When choosing axes about which to calculate torque, we can be clever and make the problem easy....
Physics 1501: Lecture 25, Pg 3
Lecture 25, Lecture 25, Act 1Act 1StaticsStatics
A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case.In which cases does the box tip over ?
A mass oscillates up & down on a spring. It’s position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration ?
t
y(t)
(a)
(b)
(c)
Physics 1501: Lecture 25, Pg 18
ExampleExample A mass m = 2kg on a spring oscillates with amplitude A =
10cm. At t=0 its speed is maximum, and is v = +2 m/s.What is the angular frequency of oscillation ?What is the spring constant k ?
A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction):
k
m
y
0
d
(a) v(t) = - vmax sin(t) a(t) = -amax cos(t)
(b) v(t) = vmax sin(t) a(t) = amax cos(t)
(c) v(t) = vmax cos(t) a(t) = -amax cos(t)
(both vmax and amax are positive numbers)
t = 0
Physics 1501: Lecture 25, Pg 22
Energy of the Spring-Mass SystemEnergy of the Spring-Mass System
We know enough to discuss the mechanical energy of the oscillating mass on a spring.
Kinetic energy is always K = 1/2 mv2
K = 1/2 m (-Asin(t + ))2
We also know what the potential energy of a spring is,
U = 1/2 k x2
U = 1/2 k (Acos(t + ))2
x(t) = Acos(t + )
v(t) = -Asin(t + )
a(t) = -2Acos(t + )
Remember,
Physics 1501: Lecture 25, Pg 23
Energy of the Spring-Mass SystemEnergy of the Spring-Mass System
Add to get E = K + U
1/2 m (A)2sin2(t + ) + 1/2 k (Acos(t + ))2
Remember that
U~cos2K~sin2
E = 1/2 kA2
so, E = 1/2 kA2 sin2(t + ) + 1/2 kA2 cos2(t + )
= 1/2 kA2 [ sin2(t + ) + cos2(t + )]
= 1/2 kA2
Physics 1501: Lecture 25, Pg 24
SHM So FarSHM So Far
The most general solution is x = Acos(t + )
where A = amplitude
= frequency
= phase constant
For a mass on a spring
The frequency does not depend on the amplitude !!!We will see that this is true of all simple harmonic
motion ! The oscillation occurs around the equilibrium point where
the force is zero!
Physics 1501: Lecture 25, Pg 25
The Simple PendulumThe Simple Pendulum
A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small
displacements.
L
m
mg
z
Physics 1501: Lecture 25, Pg 26
The Simple Pendulum...The Simple Pendulum...
Recall that the torque due to gravity about the rotation (z) axis is = -mgd.
d = Lsin Lfor small
so = -mg L
L
dm
mg
z
where
Differential equation for simple harmonic motion !