An object moving along the x-axis is said to exhibit simple harmonic motion if its position as a function of time varies as
x(t) = x0 +
Acos(ωt + φ).
The object oscillates about the equilibrium position x0. If we choose the
origin of our coordinate system such that x0 = 0, then the displacement x from
the equilibrium position as a function of time is given by
x(t) = Acos(ωt +
φ).
A is the amplitude of the oscillation, i.e. the
maximum displacement of the object from equilibrium, either in the positive or negative
x-direction. Simple harmonic motion is repetitive. The period T
is the time it takes the object to complete one oscillation and return to the starting
position. The angular frequency ω
is given by ω = 2π/T. The angular
frequency is measured in radians per second. The inverse of the period is the frequency
f = 1/T.
The frequency f = 1/T = ω/2π of the motion gives the number of complete oscillations per unit
time. It is measured in units of Hertz, (1Hz = 1/s).The velocity of the object as a function of time is given by
v(t) = dx(t)/dt = -ωAsin(ωt +
φ),
and the acceleration is given by
a(t) = dv(t)/dt = -ω2Acos(ωt
+ φ)
= -ω2x.
The quantity φ is called the phase
constant. It is determined by the initial conditions of the
motion.
If at t = 0
the object has its maximum displacement in the positive x-direction,
then φ = 0, if it has its maximum displacement in the negative
x-direction,
then φ = π. If at t = 0 the particle is moving through its
equilibrium position with maximum velocity in the negative x-direction
then φ
= π/2.
The quantity ωt + φ
is called the phase.In the figure below position and velocity are plotted as a function of time for oscillatory motion with a period of 5 s. The amplitude and the maximum velocity have arbitrary units. Position and velocity are out of phase. The velocity is zero at maximum displacement, and the displacement is zero at maximum speed.
F = ma = -mω2x .
It obeys Hooke's law, F = -kx, with k = mω2.F = ma = md2x/dt2 with F = -kx, leads to the second-order differential equation
d2x/dt2 = -(k/m)x.
We now know how to solve this equation. The solution is
x(t) = Acos(ωt +
φ),
with ω2 = k/m.
The solution to a second-order differential equation includes two constants of
integration. Here these constants are A and φ.
They are
determined by the initial conditions of the problem.The same equations have the same solutions. Whenever you encounter a differential equation of the form d2x/dt2 = -b2x, you know that the solution is x(t) = Acos(ωt + φ), with ω = b.
F = -kx
Assume the spring is stretched a distance A and then released. The object attached to
the spring accelerates.
a = -(k/m)x
It gains speed as it moves towards the equilibrium position because its acceleration is
in the direction of its velocity. When it is at the equilibrium position, the acceleration
is zero, but the object has kinetic energy. It overshoots the equilibrium position and
starts slowing down, because the acceleration is now in a direction opposite to the
direction of its velocity. Neglecting friction, it comes to a stop when the spring is
compressed by a distance A and then accelerates back towards the equilibrium position.
It
again overshoots and comes to a stop at the initial position when the spring is stretched
a distance A. The motion repeats. The object oscillates back and forth.
It executes simple
harmonic motion. The angular frequency of the motion is 
, 
, 
.Problems:
 | A particle oscillates with simple harmonic motion, so that its
displacement varies according to the expression x = (5
cm)cos(2t + π/6)
where x is in centimeters and t is in seconds. At t = 0 find (a) the displacement of the particle, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion.
| ||||||||||
| A 20 g particle moves in simple harmonic motion with a frequency of 3
oscillations per second and an amplitude of 5cm. (a) Through what total distance does the particle move during one cycle of its motion? (b) What is its maximum speed? Where does that occur? (c) Find the maximum acceleration of the particle. Where in the motion does the maximum acceleration occur?
| ||||||||||
| A 1kg mass attached to a spring of force constant 25N/m oscillates on a
horizontal frictionless track. At t = 0 the mass is released from rest
at x = -3cm, that is the spring is compressed by 3cm. Neglect the mass
of the spring. Find (a) The period of its motion, (b) the maximum value of its speed and acceleration, and (c) the displacement, velocity and acceleration as a function of time.
|
What is the total energy of the object?
The object's kinetic energy is
K = (1/2)mv2 = (1/2)mω2A2sin2(ωt
+ φ),
Its potential energy is elastic potential energy. The elastic potential energy stored
in a spring displaced a distance x from its equilibrium position is U=(1/2)kx2.
The object's potential energy therefore is
U = (1/2)kx2 = (1/2)mω2x2
= (1/2)mω2A2cos2(ωt
+ φ).
The total mechanical energy of the object is
E = K+U = (1/2)mω2A2(sin2(ωt
+ φ)+cos2(ωt
+ φ))
= (1/2)mω2A2.
The energy E in the system is proportional to the square of the amplitude.
E = (1/2)kA2.
It is a continuously changing mixture of kinetic energy and potential energy. For any object executing simple harmonic motion with angular frequency w, the restoring force F = -mω2x obeys Hooke's law, and therefore is a conservative force. We can define a potential energy U = (1/2)mω2x 2, and the total energy of the object is given by E = (1/2)mω2A2.
Problems:
| A particle that hangs from a spring oscillates with an angular frequency
of 2 rad/s. The spring is suspended from the ceiling of an elevator
car and hangs motionless (relative to the car) as the car descends at a
constant speed of 1.5 m/s. The car then suddenly stops. Neglect
the mass of the spring. (a) With what amplitude does the particle oscillate? (b) What is the equation of motion for the particle? (Choose the upward direction to be positive.)
|
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