harmonic oscillator
{{merge|Simple harmonic motion|Talk:Harmonic oscillator#Merger proposal|date=September 2008}}{{Cleanup|date=June 2008}}
missing image!
- Simple harmonic oscillator.gif -
frame|An undamped spring-mass system is a simple harmonic oscillator.
In
classical mechanics, a
harmonic oscillator is a system which, when displaced from its equilibrium position, experiences a restoring
force F
proportional to the displacement
x
according to
Hooke's law:
where
k
is a positive
constant.If
F
is the only force acting on the system, the system is called a
simple harmonic oscillator, and it undergoes
simple harmonic motion:
sinusoidal oscillations about the equilibrium point, with a constant
amplitude and a constant
frequency (which does not depend on the
amplitude).If a frictional
force (
damping) proportional to the
velocity is also present, the harmonic oscillator is described as a
damped oscillator. In such situation, the
frequency of the
oscillations is smaller than in the non-damped case, and the
amplitude of the
oscillations decreases with time.If an external time dependent
force is present, the harmonic oscillator is described as a
driven oscillator.Mechanical examples include
pendula (with small angles of displacement), masses connected to
springs, and
acoustical systems. Other analogous systems include electrical harmonic oscillators such as
RLC circuits (see
Equivalent systems below). The analysis of the harmonic oscillator exhibits the paradox that there is almost no ideal, or perfect, harmonic oscillator in nature or by manufacture, yet it is a system of profound importance in mathematics, physics and applied science.
Simple harmonic oscillator
The simple harmonic oscillator has no driving force, and no
friction (
damping), so the net force is just:
Using
Newton's Second Law of motion,
The acceleration,
a
is equal to the second derivative of
x
.
m mathrmdarg∈-→(:-4(x;font-size:12(x;">2x/mathrmdtarg∈-→(:-4(x;font-size:12(x;">2 = -k x
If we define
ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2 = k/m
, then the equation can be written as follows,
mathrmdarg∈-→(:-4(x;font-size:12(x;">2x/mathrmdtarg∈-→(:-4(x;font-size:12(x;">2 + ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2 x = 0
Define
deriv(⋅) x = mathrmdx/mathrmdt
.
We observe that:
mathrmdarg∈-→(:-4(x;font-size:12(x;">2 x/mathrmd targ∈-→(:-4(x;font-size:12(x;">2 = dderiv(⋅) x = mathrmdderiv(⋅) x/mathrmdtmathrmdx/mathrmdx=mathrmdderiv(⋅) x/mathrmdxmathrmdx/mathrmdt=mathrmdderiv(⋅)x/mathrmdxderiv(⋅) x
and substituting
mathrmd deriv(⋅)x/mathrmdxderiv(⋅) x + ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2 x = 0
mathrmd deriv(⋅)xcderiv(⋅) deriv(⋅) x + ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2 x cderiv(⋅) mathrmdx = 0
integrating
deriv(⋅)xarg∈-→(:-4(x;font-size:12(x;">2 + ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2 xarg∈-→(:-4(x;font-size:12(x;">2 = K
where
K is the
integration constant, set
K = (
A ω0)
2
deriv(⋅)xarg∈-→(:-4(x;font-size:12(x;">2 = Aarg∈-→(:-4(x;font-size:12(x;">2 ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2-ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2 xarg∈-→(:-4(x;font-size:12(x;">2
deriv(⋅)x = &(lusmn; ωarg∈-→(:4(x;font-size:12(x;">0 √Aarg∈-→(:-4(x;font-size:12(x;">2 - xarg∈-→(:-4(x;font-size:12(x;">2
frac mathrmdx/&(lusmn; √Aarg∈-→(:-4(x;font-size:12(x;">2 - xarg∈-→(:-4(x;font-size:12(x;">2 = ωarg∈-→(:4(x;font-size:12(x;">0mathrmdt
integrating, the results (including integration constant ϕ) are
beg∈cases arcs∈frac x/A= ωarg∈-→(:4(x;font-size:12(x;">0 t + &(hi; arccosfrac x/A= ωarg∈-→(:4(x;font-size:12(x;">0 t + &(hi; endcases
and has the general solution
x = A cos (ωarg∈-→(:4(x;font-size:12(x;">0 t + &(hi;)
where the
amplitude A
and the
phase &(hi;
are determined by the initial conditions.Alternatively, the general solution can be written as
x = A s∈ (ωarg∈-→(:4(x;font-size:12(x;">0 t + &(hi;)
where the value of
&(hi;
is shifted by
&(i;/2
relative to the previous form;or as
x = Carg∈-→(:4(x;font-size:12(x;">1 s∈ωarg∈-→(:4(x;font-size:12(x;">0 t + Carg∈-→(:4(x;font-size:12(x;">2 cosωarg∈-→(:4(x;font-size:12(x;">0 t
where
Carg∈-→(:4(x;font-size:12(x;">1
and
Carg∈-→(:4(x;font-size:12(x;">2
are the constants which are determined by the initial conditions.The
frequency of the oscillations is given by
displaystyle
f
=
frac
{omega_0}
{2 pi}
=
frac{1}{2 pi}
sqrt{frac{k}{m}}
The
kinetic energy is
K = 1/2 m ((mathrmdx/mathrmdt&nbs(;))arg∈-→(:-4(x;font-size:12(x;">2 = 1/2 k Aarg∈-→(:-4(x;font-size:12(x;">2 s∈arg∈-→(:-4(x;font-size:12(x;">2(ωarg∈-→(:4(x;font-size:12(x;">0 t + &(hi;)
.
and the
potential energy is
U = 1/2 k xarg∈-→(:-4(x;font-size:12(x;">2 = 1/2 k Aarg∈-→(:-4(x;font-size:12(x;">2 cosarg∈-→(:-4(x;font-size:12(x;">2(ωarg∈-→(:4(x;font-size:12(x;">0 t + &(hi;)
so the
total energy of the system has the constant value
E = 1/2 k Aarg∈-→(:-4(x;font-size:12(x;">2.
Driven harmonic oscillator
A driven harmonic oscillator satisfies the nonhomogeneous second order
linear differential equation
mathrmdarg∈-→(:-4(x;font-size:12(x;">2x/mathrmdtarg∈-→(:-4(x;font-size:12(x;">2 + ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2x = Aarg∈-→(:4(x;font-size:12(x;">0 cos(ω t)
where
Aarg∈-→(:4(x;font-size:12(x;">0
is the driving amplitude and
ω
is the driving
frequency for a sinusoidal driving mechanism. This type of system appears in
AC LC (
inductor-
capacitor) circuits and idealized spring systems lacking internal mechanical resistance or external
air resistance.
Damped harmonic oscillator
A damped harmonic oscillator satisfies the second order differential equation
mathrmdarg∈-→(:-4(x;font-size:12(x;">2x/mathrmdtarg∈-→(:-4(x;font-size:12(x;">2 + b/m mathrmdx/mathrmdt + ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2x = 0
where
b
is an experimentally determined damping constant satisfying the relationship
F = -bv
. An example of a system obeying this equation would be a weighted spring underwater if the damping force exerted by the water is assumed to be linearly proportional to
v
.The frequency of the damped harmonic oscillator is given by
ωarg∈-→(:4(x;font-size:12(x;">1 = √ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2 - Rarg∈-→(:4(x;font-size:12(x;">marg∈-→(:-4(x;font-size:12(x;">2
where
Rarg∈-→(:4(x;font-size:12(x;">m=b/2m.
Damped, driven harmonic oscillator
This satisfies the equation
mmathrmdarg∈-→(:-4(x;font-size:12(x;">2x/mathrmdtarg∈-→(:-4(x;font-size:12(x;">2 + r mathrmdx/mathrmdt + kx= Farg∈-→(:4(x;font-size:12(x;">0 cos(ω t).
The general solution is a sum of a
transient (the solution for damped undriven harmonic oscillator,
homogeneous ODE) that depends on initial conditions, and a
steady state (particular solution of the nonhomogenous ODE) that is independent of initial conditions and depends only on driving frequency, driving force, restoring force, damping force, The steady-state solution is
x(t) = Farg∈-→(:4(x;font-size:12(x;">0/Zarg∈-→(:4(x;font-size:12(x;">m ω s∈(ω t - &(hi;)
where
Zarg∈-→(:4(x;font-size:12(x;">m = √rarg∈-→(:-4(x;font-size:12(x;">2 + ((ω m - k/ω&nbs(;))arg∈-→(:-4(x;font-size:12(x;">2
is the absolute value of the
impedance or
linear response function
Z = r + i((ω m - k/ω&nbs(;))
and
&(hi; = arctan((ω m - k/ω/r&nbs(;))
is the
phase of the oscillation relative to the driving force. One might see that for a certain driving frequency,
ω
, the amplitude (relative to a given
Farg∈-→(:4(x;font-size:12(x;">0
) is maximal. This occurs for the frequency
ωarg∈-→(:4(x;font-size:12(x;">r = √k/m - 2((r/2 m&nbs(;))arg∈-→(:-4(x;font-size:12(x;">2
and is called
resonance of displacement.In summary: at a steady state the frequency of the oscillation is the same as that of the driving force, but the oscillation is phase-offset and scaled by amounts that depend on the frequency of the driving force in relation to the preferred (resonant) frequency of the oscillating system.Example:
RLC circuit.
Full mathematical definition
Most harmonic oscillators, at least approximately, solve the differential equation:
mathrmdarg∈-→(:-4(x;font-size:12(x;">2x/mathrmdtarg∈-→(:-4(x;font-size:12(x;">2 + b/m mathrmdx/mathrmdt + ωarg∈-→(:4(x;font-size:12(x;">0arg∈-→(:-4(x;font-size:12(x;">2x = Aarg∈-→(:4(x;font-size:12(x;">0 cos(ω t)
where
t is time,
b is the damping constant, ω
o is the characteristic
angular frequency, and
Aocos(ω
t) represents something driving the system with amplitude
Ao and angular frequency ω.
x is the measurement that is oscillating; it can be position, current, or nearly anything else. The
angular frequency is related to the frequency,
f, by
Important terms
- Amplitude: maximal displacement from the equilibrium.
- Period: the time it takes the system to complete an oscillation cycle. Inverse of frequency.
- Frequency: the number of cycles the system performs per unit time (usually measured in hertz = 1/s).
- Angular frequency:
ω = 2 &(i; f
- Phase: how much of a cycle the system completed (system that begins is in phase zero, system which completed half a cycle is in phase
&(i;
).
- Initial conditions: the state of the system at t = 0, the beginning of oscillations.
Universal oscillator equation
The equation
mathrmdarg∈-→(:-4(x;font-size:12(x;">2q/mathrmd τarg∈-→(:-4(x;font-size:12(x;">2 + 2 &zη; mathrmdq/mathrmdτ + q = 0
is known as the
universal oscillator equation since all second order linear oscillatory systems can be reduced to this form. This is done through
nondimensionalization.If the forcing function is
f(
t) = cos(
ωt) = cos(
ωtcτ) = cos(ω
τ), where ω =
ωtc, the equation becomes
mathrmdarg∈-→(:-4(x;font-size:12(x;">2q/mathrmd τarg∈-→(:-4(x;font-size:12(x;">2 + 2 &zη; mathrmdq/mathrmdτ + q = cos(ω τ).
The solution to this differential equation contains two parts, the "transient" and the "steady state".
Transient solution
The solution based on solving the
ordinary differential equation is for arbitrary constants
c1 and
c2 is
qarg∈-→(:4(x;font-size:12(x;">t (τ) = beg∈cases earg∈-→(:-4(x;font-size:12(x;">-&zη;τ (( carg∈-→(:4(x;font-size:12(x;">1 earg∈-→(:-4(x;font-size:12(x;">τ √&zη;arg∈-→(:-4(x;font-size:12(x;">2 - 1 + carg∈-→(:4(x;font-size:12(x;">2 earg∈-→(:-4(x;font-size:12(x;">- τ √&zη;arg∈-→(:-4(x;font-size:12(x;">2 - 1 &nbs(;)) & &zη; > 1 (/dam&(i;ng) earg∈-→(:-4(x;font-size:12(x;">-&zη;τ (carg∈-→(:4(x;font-size:12(x;">1+carg∈-→(:4(x;font-size:12(x;">2 τ) = earg∈-→(:-4(x;font-size:12(x;">-τ(carg∈-→(:4(x;font-size:12(x;">1+carg∈-→(:4(x;font-size:12(x;">2 τ) & &zη; = 1 (critical dam&(i;ng) earg∈-→(:-4(x;font-size:12(x;">-&zη; τ ([ carg∈-→(:4(x;font-size:12(x;">1 cos ((√1-&zη;arg∈-→(:-4(x;font-size:12(x;">2 τ&nbs(;)) +carg∈-→(:4(x;font-size:12(x;">2 s∈((√1-&zη;arg∈-→(:-4(x;font-size:12(x;">2 τ&nbs(;)) &nbs(;)] & &zη; < 1 (underdam&(i;ng) endcases
The transient solution is independent of the forcing function. If the system is critically damped, the response is independent of the damping.
Steady-state solution
Apply the "
complex variables method" by solving the
auxiliary equation below and then finding the real part of its solution:
mathrmdarg∈-→(:-4(x;font-size:12(x;">2 q/mathrmdτarg∈-→(:-4(x;font-size:12(x;">2 + 2 &zη; mathrmdq/mathrmdτ + q = cos(ω τ) + is∈(ω τ) = earg∈-→(:-4(x;font-size:12(x;"> i ω τ .
Supposing the solution is of the form
qarg∈-→(:4(x;font-size:12(x;">s(τ) = A earg∈-→(:-4(x;font-size:12(x;">i ( ω τ + &(hi; ) .
Its derivatives from zero to 2nd order are
qarg∈-→(:4(x;font-size:12(x;">s = A earg∈-→(:-4(x;font-size:12(x;">i ( ω τ + &(hi; ) mathrmdqarg∈-→(:4(x;font-size:12(x;">s/mathrmd τ = i ω A earg∈-→(:-4(x;font-size:12(x;">i ( ω τ + &(hi; ) mathrmdarg∈-→(:-4(x;font-size:12(x;">2 qarg∈-→(:4(x;font-size:12(x;">s/mathrmd τarg∈-→(:-4(x;font-size:12(x;">2 = - ωarg∈-→(:-4(x;font-size:12(x;">2 A earg∈-→(:-4(x;font-size:12(x;">i ( ω τ + &(hi; ) .
Substituting these quantities into the differential equation gives
-ωarg∈-→(:-4(x;font-size:12(x;">2 A earg∈-→(:-4(x;font-size:12(x;">i (ω τ + &(hi;) + 2 &zη; i ω A earg∈-→(:-4(x;font-size:12(x;">i(ω τ + &(hi;) + A earg∈-→(:-4(x;font-size:12(x;">i(ω τ + &(hi;) = (-ωarg∈-→(:-4(x;font-size:12(x;">2 A + 2 &zη; i ω A + A) earg∈-→(:-4(x;font-size:12(x;">i (ω τ + &(hi;) = earg∈-→(:-4(x;font-size:12(x;">i ω τ .
Dividing by the exponential term on the left results in
-ωarg∈-→(:-4(x;font-size:12(x;">2 A + 2 &zη; i ω A + A = earg∈-→(:-4(x;font-size:12(x;">-i &(hi; = cos&(hi; - i s∈&(hi; .
Equating the real and imaginary parts results in two independent equations
A (1-ωarg∈-→(:-4(x;font-size:12(x;">2)=cos&(hi; &nbs(;&nbs(;&nbs(;&nbs(; 2 &zη; ω A = - s∈&(hi;.
Amplitude part
thumb|Bode plot of the frequency response of an ideal harmonic oscillator.Squaring both equations and adding them together gives
(&nbs(;. matrixAarg∈-→(:-4(x;font-size:12(x;">2 (1-ωarg∈-→(:-4(x;font-size:12(x;">2)arg∈-→(:-4(x;font-size:12(x;">2 = cosarg∈-→(:-4(x;font-size:12(x;">2&(hi; (2 &zη; ω A)arg∈-→(:-4(x;font-size:12(x;">2 = s∈arg∈-→(:-4(x;font-size:12(x;">2&(hi; endmatrix &nbs(;) ⇒ Aarg∈-→(:-4(x;font-size:12(x;">2[(1-ωarg∈-→(:-4(x;font-size:12(x;">2)arg∈-→(:-4(x;font-size:12(x;">2 + (2 &zη; ω)arg∈-→(:-4(x;font-size:12(x;">2] = 1.
By convention the positive root is taken since amplitude is usually considered a positive quantity. Therefore,
A = A( &zη; ω) = 1/√(1-ωarg∈-→(:-4(x;font-size:12(x;">2)arg∈-→(:-4(x;font-size:12(x;">2 + (2 &zη; ω)arg∈-→(:-4(x;font-size:12(x;">2.
Compare this result with the theory section on
resonance, as well as the "magnitude part" of the
RLC circuit. This amplitude function is particularly important in the analysis and understanding of the
frequency response of second-order systems.
Phase part
To solve for φ, divide both equations to get
tan&(hi; = - 2 &zη; ω/ 1 - ωarg∈-→(:-4(x;font-size:12(x;">2 = 2 &zη; ω/ωarg∈-→(:-4(x;font-size:12(x;">2 - 1 ⇒ &(hi; ≡ &(hi;(&zη; ω) = arctan (( 2 &zη; ω/ωarg∈-→(:-4(x;font-size:12(x;">2 - 1 &nbs(;) ).
This phase function is particularly important in the analysis and understanding of the
frequency response of second-order systems.
Full solution
Combining the amplitude and phase portions results in the steady-state solution
qarg∈-→(:4(x;font-size:12(x;">s (τ) = A(&zη;ω) cos(ω τ + &(hi;(&zη;ω)) = Acos(ω τ + &(hi;).
The solution of original universal oscillator equation is a
superposition (sum) of the transient and steady-state solutions
q(τ) = qarg∈-→(:4(x;font-size:12(x;">t (τ) + qarg∈-→(:4(x;font-size:12(x;">s (τ).
For a more complete description of how to solve the above equation, see
linear ODEs with constant coefficients.
Equivalent systems
Harmonic oscillators occurring in a number of areas of engineering are equivalent in the sense that their mathematical models are identical (see
universal oscillator equation above). Below is a table showing analogous quantities in four harmonic oscillator systems in mechanics and electronics. If analogous parameters on the same line in the table are given numerically equal values, the behavior of the oscillators will be the same. {|class="wikitable" cellpadding="4" style="background:#F8F8F8;"!width="225" align="left"|Translational Mechanical!width="225" align="left"|Torsional Mechanical!width="225" align="left"|Series RLC Circuit!width="225" align="left"|Parallel RLC Circuit
|
x | > thη | >Charge q | >|Voltage e
|
| Velocity dx/dt | >Angular velocity dthη/dt | >Current dq/dt | >|de/dt
|
| Mass M | >Moment of inertia I | >Inductance L | >|Capacitance C
|
| Hooke's law>Spring constant | K
Torsion spring#Torsion coefficient | μ | >Capacitance#Elastance>Elastance | 1/C
Susceptance | 1/L
|
| Friction γ | >Torsion spring#Motion of torsion balances and pendulums>Rotational friction | Γ
Electrical resistance | R | >|Conductance 1/R
|
| force F(t) | >torque τ(t) | >e | >|di/dt
|
| Undamped resonant frequency | farg∈-→(:4(x;font-size:12(x;">n
:
|
1/2&(i;√K/M | >1/2&(i;√μ/I | >1/2&(i;√1/LC | >|1/2&(i;√1/LC
|
| Differential equation: |
|
|Mdderiv(⋅) x + γderiv(⋅) x + Kx = F
||Idderiv(⋅) thη + Γderiv(⋅) thη + μ thη = τ
||Ldderiv(⋅) q + Rderiv(⋅) q + q/C = e
||Cdderiv(⋅) e + deriv(⋅) e/R + e/L = deriv(⋅) i
Applications
The problem of the simple harmonic oscillator occurs frequently in physics because a mass at equilibrium under the influence of any conservative force, in the limit of small motions, will behave as a simple harmonic oscillator. A conservative force is one that has a potential energy function. The potential energy function of a harmonic oscillator is:
V(x) = 1/2 k xarg∈-→(:-4(x;font-size:12(x;">2
Given an arbitrary potential energy function V(x)
, one can do a Taylor expansion in terms of x
around an energy minimum (x = xarg∈-→(:4(x;font-size:12(x;">0
) to model the behavior of small perturbations from equilibrium.
V(x) = V(xarg∈-→(:4(x;font-size:12(x;">0) + (x-xarg∈-→(:4(x;font-size:12(x;">0) V'(xarg∈-→(:4(x;font-size:12(x;">0) + 1/2 (x-xarg∈-→(:4(x;font-size:12(x;">0)arg∈-→(:-4(x;font-size:12(x;">2 Varg∈-→(:-4(x;font-size:12(x;">(2)(xarg∈-→(:4(x;font-size:12(x;">0) + O(x-xarg∈-→(:4(x;font-size:12(x;">0)arg∈-→(:-4(x;font-size:12(x;">3
Because V(xarg∈-→(:4(x;font-size:12(x;">0)
is a minimum, the first derivative evaluated at xarg∈-→(:4(x;font-size:12(x;">0
must be zero, so the linear term drops out:
V(x) = V(xarg∈-→(:4(x;font-size:12(x;">0) + 1/2 (x-xarg∈-→(:4(x;font-size:12(x;">0)arg∈-→(:-4(x;font-size:12(x;">2 Varg∈-→(:-4(x;font-size:12(x;">(2)(xarg∈-→(:4(x;font-size:12(x;">0) + O(x-xarg∈-→(:4(x;font-size:12(x;">0)arg∈-→(:-4(x;font-size:12(x;">3
The constant term V(x0) is arbitrary and thus may be dropped, and a coordinate transformation allows the form of the simple harmonic oscillator to be retrieved:
V(x) &asy&(lusmn;; 1/2 xarg∈-→(:-4(x;font-size:12(x;">2 Varg∈-→(:-4(x;font-size:12(x;">(2)(0) = 1/2 k xarg∈-→(:-4(x;font-size:12(x;">2
Thus, given an arbitrary potential energy function V(x)
with a non-vanishing second derivative, one can use the solution to the simple harmonic oscillator to provide an approximate solution for small perturbations around the equilibrium point.Examples
Simple pendulum
missing image!
- Simple pendulum height.png -
A simple pendulum exhibits simple harmonic motion under the conditions of no damping and small amplitude.
Assuming no damping and small amplitudes, the differential equation governing a simple pendulum is
mathrmdarg∈-→(:-4(x;font-size:12(x;">2thη/ mathrmdtarg∈-→(:-4(x;font-size:12(x;">2+g/ ellthη=0.
The solution to this equation is given by:
thη(t) = thηarg∈-→(:4(x;font-size:12(x;">0cos((√g/ ellt&nbs(;)) &nbs(;&nbs(; &nbs(;&nbs(; &nbs(;&nbs(; &nbs(;&nbs(; ||thηarg∈-→(:4(x;font-size:12(x;">0|| ll 1
where thηarg∈-→(:4(x;font-size:12(x;">0
is the largest angle attained by the pendulum. The period, the time for one complete oscillation , is given by 2&(i;
divided by whatever is multiplying the time in the argument of the cosine (√g/ ell
here).
Targ∈-→(:4(x;font-size:12(x;">0 = 2&(i;√ell/ g &nbs(;&nbs(; &nbs(;&nbs(; &nbs(;&nbs(; &nbs(;&nbs(; ||thηarg∈-→(:4(x;font-size:12(x;">0|| ll 1.
Pendulum swinging over turntable
Simple harmonic motion can in some cases be considered to be the one-dimensional projection of two-dimensional circular motion. Consider a long pendulum swinging over the turntable of a record player. On the edge of the turntable there is an object. If the object is viewed from the same level as the turntable, a projection of the motion of the object seems to be moving backwards and forwards on a straight line. It is possible to change the frequency of rotation of the turntable in order to have a perfect synchronization with the motion of the pendulum.The angular speed of the turntable is the pulsation of the pendulum.In general, the pulsation-also known as angular frequency, of a straight-line simple harmonic motion is the angular speed of the corresponding circular motion.Therefore, a motion with period T and frequency f=1/T has pulsation
In general, pulsation and angular speed are not synonymous. For instance the pulsation of a pendulum is not the angular speed of the pendulum itself, but it is the angular speed of the corresponding circular motion.Spring-mass system
thumb|right|250px|Spring-mass system in equilibrium (A), compressed (B) and stretched (C) states. When a spring is stretched or compressed by a mass, the spring develops a restoring force. Hooke's law gives the relationship of the force exerted by the spring when the spring is compressed or stretched a certain length:
F (( t &nbs(;)) =-kx (( t &nbs(;))
where F is the force, k is the spring constant, and x is the displacement of the mass with respect to the equilibrium position.This relationship shows that the distance of the spring is always opposite to the force of the spring.By using either force balance or an energy method, it can be readily shown that the motion of this system is given by the following differential equation:
F(t) = -kx(t) = m frac mathrmdarg∈-→(:-4(x;font-size:12(x;">2/mathrmdarg∈-→(:-4(x;font-size:12(x;">2 x (( t &nbs(;)) = ma.
...the latter evidently being (Newton%27s_second_law_of_motion#Newton.27s_second_law:_law_of_resultant_force|Newton's second law of motion).If the initial displacement is A, and there is no initial velocity, the solution of this equation is given by:
x (( t &nbs(;)) =Acos (( (√ k/m) t&nbs(;)).
Energy variation in the spring-damper system
In terms of energy, all systems have two types of energy, potential energy and kinetic energy. When a spring is stretched or compressed, it stores elastic potential energy, which then is transferred into kinetic energy. The potential energy within a spring is determined by the equation U = 1/2kxarg∈-→(:-4(x;font-size:12(x;">2.
When the spring is stretched or compressed, kinetic energy of the mass gets converted into potential energy of the spring. By conservation of energy, assuming the datum is defined at the equilibrium position, when the spring reaches its maximum potential energy, the kinetic energy of the mass is zero. When the spring is released, the spring will try to reach back to equilibrium, and all its potential energy is converted into kinetic energy of the mass.References
- BOOK, Serway, Raymond A., Jewett, John W., Physics for Scientists and Engineers, Brooks/Cole, 2003, 0-534-40842-7,
- BOOK, Tipler, Paul, Physics for Scientists and Engineers: Vol. 1, 4th ed., W. H. Freeman, 1998, 1-57259-492-6,
- BOOK, Wylie, C. R., Advanced Engineering Mathematics, 4th ed., McGraw-Hill, 1975, 0-07-072180-7,
See also
External links
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