Chapter 11humanic/p1200_lecture22… · · 2018-03-30Chapter 11 Vibrations and Waves . The Ideal...
Transcript of Chapter 11humanic/p1200_lecture22… · · 2018-03-30Chapter 11 Vibrations and Waves . The Ideal...
Chapter 11
Vibrations and Waves
The Ideal Spring and Simple Harmonic Motion
xkF Appliedx =
spring constant Units: N/m
If you apply a force FxApplied to an ideal spring, it will compress or
expand by an amount x according to the relation,
The Ideal Spring and Simple Harmonic Motion
HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING The restoring force on an ideal spring is xkFx −=
Simple harmonic motion is a type of periodic motion where the restoring force on a mass is directly proportional to the displacement. The displacement of the mass has a sinusoidal dependence on the elapsed time, i.e. depends on time through a sine or cosine function.
A mass-spring system undergoing Simple Harmonic Motion
A mass-spring system undergoing Simple Harmonic Motion and tracing out a sinusoidal function of its displacement as time elapses
The Ideal Spring and Simple Harmonic Motion
Energy and Simple Harmonic Motion
DEFINITION OF ELASTIC POTENTIAL ENERGY The elastic potential energy is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring, the elastic potential energy is
221
elasticPE kx=
SI Unit of Elastic Potential Energy: joule (J)
Since the restoring force due to an ideal spring is a conservative force, the total mechanical energy of a spring-mass system is conserved, and is
Energy and Simple Harmonic Motion
E = KE +PE = 12 mv
2 + 12 kx
2
Since v = 0 at ± xmax ⇒ E = 12 kxmax
2
∴ 12 kxmax
2 = 12 mv
2 + 12 kx
2
⇒ v = ± km
xmax2 − x2
v = ±vmax = ±xmaxkm
at x = 0
Simple Harmonic Motion and the Reference Circle
How do x, v and a depend on time quantitatively in SHM? Since F=-kx does not give constant acceleration à can’t used our constant acceleration equations. As an analogy with the sinusoidal dependence of displacement on time in SHM, consider the projection on the x-axis of the position of a ball undergoing uniform circular motion on a turntable. Record how x depends on time with a moving roll of film.
Simple Harmonic Motion and the Reference Circle
Since θ =ωt,⇒ x = Acosθ = Acosωt
Assume that the turntable is rotating at the constant angular speed ω andthe ball is at radius A.
DISPLACEMENT
Simple Harmonic Motion and the Reference Circle
period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
Tf 1= ω =
2πT
= 2π f
amplitude A: the maximum displacement
Simple Harmonic Motion and the Reference Circle
VELOCITY
vx = −vT sinθ = −Aωvmax!sinωt
Simple Harmonic Motion and the Reference Circle
Example: The Maximum Speed of a Loudspeaker Diaphragm The motion of a loudspeaker is sinusoidal. The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm. (a) What is the maximum speed of the diaphragm? (b) Where in the motion does this maximum speed occur?
Simple Harmonic Motion and the Reference Circle
vx = −vT sinθ = −Aωvmax!sinωt(a)
( ) ( )( )( )sm3.1
Hz100.12m1020.02 33max
=
××=== − ππω fAAv
(b) The maximum speed occurs midway between the ends of its motion.
x = Acosωt = 0
when ωt = π2
, 3π2
,…
vx = vmax occurs when ωt = π2
, 3π2
,…
Simple Harmonic Motion and the Reference Circle
ACCELERATION
ax = −ac cosθ = −Aω2
amax!cosωt
x = Acosωt
vx = −Aωvmax!sinωt
ax = −Aω2
amax!cosωt
Displacement
Displacement, Velocity and Acceleration for SHM vs. time
Simple Harmonic Motion and the Reference Circle
A
A
A
A
Simple Harmonic Motion and the Reference Circle
Find a relationship between vx and x for the reference circle:
x = Acosωt vx = −Aω sinωt
cos2ωt + sin2ωt =1 ⇒x2
A2+
vx2
A2ω 2 =1
∴vx = ±ω A2 − x2
Compare this equation with the equation for v as a function of x that we got earlier for the spring-mass system from conservation of energy, i.e.
v = ± km
xmax2 − x2
This suggests that the sinusoidal equations of the reference circle for x, v and a are also valid for the spring-mass system if we set
ω =km
and A = xmax
Simple Harmonic Motion and the Reference Circle
Another way to extract the frequency of vibration of the spring-mass system: Use the reference circle equations and Newton’s 2nd law.
mk
=ω
tAax ωω cos2−=tAx ωcos=
xmakxF =−=∑2ωmAkA −=−
Simple Harmonic Motion and the Reference Circle
Example: A Body Mass Measurement Device The device consists of a spring-mounted chair in which the astronaut sits. The spring has a spring constant of 606 N/m and the mass of the chair is 12.0 kg. The measured period is 2.41 s. Find the mass of the astronaut.
Simple Harmonic Motion and the Reference Circle
totalmk
=ω 2total ωkm =
Tf π
πω22 ==
( ) astrochair2total 2mm
Tkm +==π
( )( )( ) kg 77.2kg 0.12
4s 41.2mN606
2
2
2
chair2astro
=−=
−=
π
πm
Tkm
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Example: Simple Harmonic Motion of a vertical spring-mass system. A 0.20 kg ball is attached to a vertical spring. The spring constant is 28 N/m. Find the equilibrium position of the mass with respect to the unstrained position of the spring and find the amplitude of the SHM about this equilibrium position after the ball is dropped from the unstrained position.
Equilibrium position:
F =ma⇒∑ kd0 −mg = 0
∴d0 =mgk=
0.20( ) 9.8( )28
= 0.070 m
Amplitude of oscillations:Ef = E0
12 kdmax
2 =mgdmax
dmax = 2mgk= 2
0.20( ) 9.8( )28
= 0.140 m
∴A = dmax − d0 = 0.140− 0.070 = 0.070 m
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dmax
} }
A A
The Pendulum
A simple pendulum consists of a mass attached to a frictionless pivot by a cable of negligible mass.
FT = −mgsinθ
For small angles: sinθ ≈θ
Since θ = sL
⇒ FT ≈ −mgLs↔ F = −kx
⇒ k↔ mgL⇒ω =
km=
mgmL
=gL
∴ω =gL
Pendulum (small angles only)
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The Pendulum
Example: Keeping Time Determine the length of a simple pendulum that will swing back and forth in simple harmonic motion with a period of 1.00 s.
22Lg
Tf ===
ππω
( ) ( ) m 248.04
sm80.9s 00.14 2
22
2
2
===ππ
gTL
2
2
4πgTL =
Damped Harmonic Motion
In simple harmonic motion, an object oscillates with a constant amplitude. In reality, friction or some other energy dissipating mechanism is always present and the amplitude decreases as time passes. This is referred to as damped harmonic motion.
Damped Harmonic Motion
1) simple harmonic motion – amplitude stays constant
2&3) underdamped – amplitude decreases but still oscillations 4) critically damped – amplitude decreases to 0 without oscillations in shortest possible time
5) Overdamped – amplitude decreases to 0 without oscillations slower than in critically damped case
Different types of damped harmonic motion: