OPENING QUESTIONS: Consider our equation for calculating the velocity of an object experiencing SHM:
v = dx/dt = -ωAsin(ωt + φ) (notice the similarity to rω)
1) Why is that term negative?
2a) What does ω represent? (2b) How is it different from our last use of ω?
3) What does φ stand for?
OBJECTIVE: I will be able to calculate the energy present for an object undergoing SHM after today's class.
WORDS/FORMULAE FOR TODAY:
- (0) F = -kx
- (1) x(t) = Acos (ωt + φ)
- (2) ω = angular frequency = (k/m)1/2 measured in radians/second
- (3) T = period = 2π/ω = 1/f
- (4) f = frequency 1/T = measured in cycles per seconds or waves/sec
- (5) v = dx/dt = -ωAsin(ωt + φ) (notice the similarity to rω)
- (6) a = dx/dt = -ω2Acos(ωt + φ)(notice the similarity to rω2)
- (7) KE = 1/2mv2 = 1/2m[-ωAsin(ωt + φ)]2 = 1/2mω2A2sin2(ωt + φ)
- (8) U = 1/2kx2 = 1/2kA2cos2(ωt + φ)]
- (9) TOTAL E = 1/2kA2
WORK O' THE DAY:
Mr Chase will lead us in a discussion of our homework
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Let's start by digesting more formulae: there are three more today!
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Recap:
Our discussions of SHM will primarily deal with one dimension (x) so we'll typically ASSUME that any motion is constrained to X.
We can accurately predict the motion of an object experiencing SHM by the following formula:
x(t) = Acos (ωt + φ)
That formula characterizes the mathematical (graphic) representation of an object experiencing SHM
- x = x direction
- (t) = with respect to time
- A = the AMPLITUDE of the resulting wave when the motion of the object is being graphed (measured in meters)
- ω = angular frequency (measured in radians/sec)
- t = time (in seconds)
- φ (spelled "phi" and pronounced "fee") = the "phase constant" or "initial phase angle" and is measured in radians
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Work through Example 15.3 using our usual approach: (sketch it first, write down initial conditions, close the book and then compare)
Work with your group to predict where potential energy in a system experiencing SHM would be greatest and least?
Now work with your groupies to predict the kinetic energy in a system experiencing SHM as to greatest and least.
Now predict the RELATIONSHIP between KE & U in such a system!
Not surprisingly, the U of an SHM system is at it's greatest when the spring is either compressed the most or at it's greatest extension.
For those tiny instances, the spring isn't moving so the KE at that point is 0
Let's take a gander:
As you can see the energy remains constant... soo
The total energy in a system experiencing SHM is:
E = 1/2 KA2
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HOMEWORK:
Problems: (Chapter 15 probs begining on page 475)
#21, 24, 29