Chapter 15 Oscillatory Motion
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Chapter 15
Oscillatory Motion
Intro
• Periodic Motion- the motion of an object that regularly repeats
• There is special case of periodic motion in which there is a force acting on an object proportional to the relative position of the object from equilibrium, and that force is directed towards equilibrium.
Intro
• A force such as this is called a restoring force, and results in Simple Harmonic Motion.
15.1 Motion of an Object Attached to a Spring
• 1st Example will be a block of mass m, attached to a spring of elastic constant k. • We’ve already studied the qualities of this system in Chapter 7, using Hooke’s Law and the work done by/on the spring.
15.1
• In this case, the spring provides the restoring force necessary for simple harmonic motion.
• If we apply Newton’s 2nd Law
kxFs
xmakx
m
kxax
15.1
• By definition, an object moves with SHM whenever its acceleration is proportional to its position and is oppositely directed to the displacement from equilibrium.
• Note: The SHM motion of a spring also applies to a vertically aligned spring AFTER the spring is allowed to stretch to its new equilibrium.
• Quick Quiz p. 454
15.2 Mathematical Representation of SHM
• Remember that acceleration is the 2nd derivative of position with respect to time.
• We denote the term k/m with the symbol ω2 which gives
xm
k
dt
xd
2
2
xdt
xd 22
2
15.2
• We’re thinking integrate this twice to get the position function, but we haven’t seen an integration where a derivative contains the original function.
• This is called a second order differential equation. To solve this, we need a function x(t) that has a 2nd derivative that is negative of the original function, multiplied by ω2.
15.2
• The cosine and sine functions do just that. – Derivative of cos = -sin– Derivative of -sin = -cos
• The full solution to this 2nd order differential is
• Where A, ω, and φ are constants representing amplitude, angular frequency and phase angle respectively.
)cos()( tAtx
15.2
• Amplitude (A) - the maximum distance from equilibrium (in the positive or negative direction).
• Angular frequency (ω) – measures how rapidly oscillations are occurring (rad/s)– For a spring
• Phase Angle (φ) – uniquely determines the position of the object at t=0.
m
k
15.2
• The term (ωt + φ) represents the phase of oscillation.
• Keep in mind, the cosine function REPEATS every time ωt increases by 2π radians.
15.2
15.2
• Quick Quizzes p. 456
• Period (T) – the time for one oscillation.
• Frequency (f) – inverse of period, the number of oscillations per second. (Hz = s-1)
2
Tk
mT 2
Tf
1 f 2
m
kf
21
15.2
• Position function
• Velocity function
• Acceleration function
)cos()( 22
2
tAdt
xdta
)sin()( tAdt
dxtv
)cos()( tAtx
15.2
• Max values (occur when trig function = 1)– Position- Amplitude– Velocity- (passing through equilibrium)
– Acceleration- (at +/- Amp)
Am
kAv max
Am
kAa 2
max
15.2
• Plots of x(t), v(t), a(t)• When x(t) is at + amp– v(t) = 0– a(t) = -amax
• When x(t) = 0– v(t) = +/- vmax
– a(t) = 0
15.2
• Quick Quizzes p. 456
• Remember, the angular frequency (ω) is determined by (k/m)1/2
• A and φ are determined by conditions at t=0
• Examples 15.1-15.3
15.3 Energy of the Simple Harmonic Oscillator
• The kinetic energy of the SHO varies with time according to
• The Elastic Potential Energy of the SHO also varies with time according to
)(sin 222212
21 tAmmvK
)(cos22212
21 tkAkxU
15.3
• The total energy is therefore
• And simplifying…
• The total mechanical energy of a SHO is a constant of the motion and is proportional to the square of the amplitude.
)(cos)(sin 22221 ttkAUKE
221 kAE
15.3
• Plots of Kinetic and Potential Energy
15.3
• From the Energy equations we can determine velocity as a function of position.
• Solve for v
• See Figure 15.11 Pg 463• Example 15.4 pg 464
2212
212
21 kAkxmvUKE
22 xAv
15.4 Comparing SHM and UCM
• Simple Harmonic motion can easily be used to drive uniform circular motion.
• Piston- steam engine (train)internal combustion engine
15.4
• Simple Harmonic Motion along a straight line can be represented by the projectionof uniform circular motionalong the diameter of areference circle.
15.4
• Uniform circular motion is often considered to be a combination of two SHM’s. – One along the x axis.– One along the y axis.– The motions are 90o
or π/2 out of phase.
Quick Quiz p 467Example 15.5
15.5 Pendulums
• Simple Pendulum exhibits periodic motion, very close to a simple harmonic motion for small angles (<10o)
• The solution to the 2nd order differential for pendulums is
• Where
)cos(max t
l
g
15.5
• The period of oscillation is given as
• The period of a simple pendulum is independent of the mass attached, solely depending on the length and gravity.
• Quick Quizzes p. 469• Example 15.6
g
lT
22
15.5
• Physical Pendulum- an object that cannot be approximated as a point mass, oscillating through an axis that does not pass through its center of mass.
• The moment of inertia of the system must be accounted for
)cos(max t
I
mgd
15.5
• The period is given as
• Example 15.7 p 470
mgd
IT
22
15.5
• Torsional Pendulum- theoscillation involves rotation,when the support wire twists.• The object is returned to equilibrium by a restoring Torque.
15.5
• Where κ (kappa) is the torsion constant, often determined by applying a known torque to twist the wire through a measurable angle θ.
• Using Newton’s 2nd Law for Torques
• Rearranged 2
2
dt
dI
Idt
d
2
2
15.5
• We see that this is the same form as our 2nd Order differential for the mass/spring system, with κ/I representing ω2.
• There is no small angle restriction for the torsional pendulum, so long as the elastic limit of the wire is not exceeded.
I
IT
2
15.6 Damped Oscillations
• So far we have discussed oscillations in ideal systems, with no loss of energy.
• In real life, energy is lost mostly to internal forms, diminishing the total ME with each oscillation.
• An oscillator whose energy diminishes with time is said to be Damped.
15.6
• One common example is when the oscillating object moves through a fluid (air), creating a resistive force as discussed in Ch 6.
• At low speeds the resistive or retarding force is proportional to the velocity as
• Where b is called the damping constant. vR b
15.6
• Applying Newton’s 2nd Law
• Again a 2nd Order Differential, the solution is a little more complicated.
• Where
2
2
dt
xdm
dt
dxbkx
)cos()( 2 tAetx tmb
2
2
m
b
m
k
15.6
• Recognize that (k/m)1/2 would be the angular
frequency in the absence of damping, called the “Natural Angular Frequency” (ωo)
• The motion occurs within the restricting envelope of the exponential decay.
22
2
m
bo
15.6
15.6
• Damping falls into three categories based on the value of b. – Underdamped- when the value of bvmax < kA• As the value of b increases the ampltitude of the
oscillations decreases rapidly.
– Critically Damped- when b reaches the critical value bc such that
• When the object is released it will approach but never pass through equilibrium
oc
m
b 2
15.6
– Overdamped- when the value of bvmax > kA and
• The fluid is so viscous that there is no oscillation, the object eventually returns to equilibrium.
oc
m
b 2
15.6
• Quick Quiz p. 472
15.7 Forced Oscillations
• Because most oscillators experience damping, an external force can be used to do positive work on the system compensating for the lossed energy.
• Often times the value of the force varies with time, but the system will often reach a steady state where the energy gained/lost is equal.
15.7
• The system will oscillate according to
• With a constant Amplitude given by)cos( tAx
2
222
mb
mFA
o
o
15.7
• Notice when ω≈ωo, the value of A will increase significantly.
• Essentially when the Forcing frequency matches the natural frequency oscillations intensify. – Examples- • Loma Prieta “World Series” Earthquake• Tacoma Narrows Bridge
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