Download Energy in Simple Harmonic Motion

Energy in Simple Harmonic Motion
As a mass on a spring goes
through its cycle of oscillation,
energy is transformed from
potential to kinetic and back to
potential.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Slide 14-12
Energy: Kinetic & Potential
Total Energy
Either in terms of
Kinetic K
(x=0)
Or Potential U
(|x|= A)
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Total energy
as
1) Kinetic energy in full
E = m(vmax)2/2
or
2) Potential energy in full
E= kA2/2
Equating these we get
(vmax)2 = kA2/m
But from UCM we
found vmax = ωA
Equating these two
expressions yieds =>
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Prob. 14-15
A block of mass m tied to a spring with spring
constant k is undergoing simple harmonic oscillation.
a)When the displacement of the mass is A/2, what fraction of the mechanical
energy is kinetic energy and what fraction of it is potential energy?
This is because total energy E = ½ kA2
b) At what displacement, as a fraction of A, is the energy half kinetic and half
potential?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
P14-16: At maximum displacement position, x = A,
a hammer hit gives the block a velocity v0 in the –x direction.
>> Use Energy Conservation (often the easiest approach)
a) Find Amplitude Α Note: At x = Α the kinetic energy is 0
kA2 = mv02 => A = (m /k) 1/2 v0
b) Find the velocity v2 at a point
where x = A/2.
2
1
⎞ 3⎛1
⎞
1 ⎛ A⎞
1
1
1
1 ⎛1 2⎞ 1
1⎛1
mv22 + k ⎜ ⎟ = mv20 + 0 J ⇒ mv 22 = mv02 − ⎜ kA ⎟ = mv02 − ⎜ mv20 ⎟ = ⎜ mv20 ⎟
2
⎠ 2
⎠ 4⎝2
⎠
2 ⎝2⎠
2
2
2
4 ⎝2
4⎝2
⇒ v2 =
3
v =
4 0
3
(0.40 m/s) = 0.346 m /s = 35 cm /s
4
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Ch. 14: 7, 15,16, 22, 28, 36 ,44, 50
P44:
Mass hanging at the end of spring in the
gravitational field of Planet X is stretched
by ΔL=31.2cm.
• Pull mass down by x=10cm and released
• Mass undergoes 10 oscillations in 14.5s
• Find g
14-P50
Ultrasound Device
m= 0.1g, f = 1MHz, max force F= 40,000 N,
• Find max amplitude A
• Find max speed v
How to convert a snapshot graph to a
history graph and vice versa?
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Two more examples: Prob. 15.7
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Prob. 15.9 Wave moving to the right at v=1m/s
Given: History graph at x=0, Asked: snapshot at t=1s
Think: same thing
happening at x=0 at t= 1s
has happened earlier
at t= 0 sec read off value and place it at x= 1
or t= -1 sec
at x= 2
And will happen at
t=2 sec read off value and place it at x=-1
! This is probably the toughest part in
this chapter. We’ll work out more
problems of this sort later
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Problem 15-8
Given snapshot graph at t=2s
Draw history graph at x=0 for t=1 s to 8 s
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Prob.15-16
Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.
Ch15-P17 snapshot graph at t=0s
Amplitude A= 4cm
Wavelength λ=12m
Frequency f = v / λ = 2.0 Hz
Ch 15- P18 History Graph at x=0m
of wave moving to the right at v=2m/s
Amplitude A = 6cm
Period T = 0.05 to 0.65 = 0.6 s
Frequency f = 1/T = 1.33 Hz
Wavelength λ = v/f = vT= 1.2 m
Ch 15-28 Intensity of EM radiation
•Sun emits EM waves with a power of 4x
1026W, what is the intensity of sunlight at
the positions of Venus, Mars and Saturn?
Intensity = Power / Area, A = 4πr2
Ch 15-P33
• •Source (loudspeaker) emits P= 35 W
• •Receiver (microphone) located r= 50 m from source.
(Microphone has area a=1cm2)
–
–
–
–
Find sound intensity, intensity level at the microphone
Intensity = Power / Area, where A= 4πr2
Intensity level: β = 10 dB log (Int / 10-12) = 90 db
Added question: What is the total power received by the
microphone? Power = Intensity x area,
• Or, simpler way: P_mic / P_total = a / A
Ch 15- 56
Given the form of
the Wave function
Can read off the
Values of A, λ, T
b) Calculate v
c) After this put
t=0.5 s, and x=0.2 m
Into the expression
for y(x,t) to get amplitude