# Problem. Simple harmonic oscillator

Document Sample

```					                                                  —1—

Problem. Simple harmonic oscillator
In this problem you will use one of the most important conservation laws in Physics, conservation
of energy to calculate the period of a simple harmonic oscillator. The ﬁgure below displays a mass
m, which is attached to a spring with a spring constant k.

x                    k
m                       unstretched

k
m                                  stretched

At t = 0, the spring is stretched to a distance A from the unstretched state and released from
rest. We then let is oscillate indeﬁnitely (assuming there is no friction involved), and measure the
displacement x(t) from the unstretched state as a function of time.We will ﬁnd the period of this
oscillatory system by solving a differential equation relating the displacement x(t) to its derivative,
i.e., the velocity of the mass m.

a) Finding the initial conditions: ﬁnd the value of the position, x(t), and velocity, dx/dt, of
the mass m at the time t = 0.
Solution. The initial displacement is A and the mass is released from rest, i.e., its velocity
is zero.                                           
dx 
x(t = 0) = A,         = 0.
dt t=0

b) Finding the conserved energy. During the motion, the potential energy of the spring (de-
1
pending on how far it is stretched from the original position) is given by 2 kx2 . The kinetic
energy of the mass depends on its velocity and is given by 1 mv 2 , where v = dx/dt. The
2
total energy in the system (mass + spring) is conserved because we assumed no friction is
involved in the problem. This total energy is the sum of kinetic and potential energies and
its conservation can be written as follows:
2
1     dx        1
m            + kx2 = constant.
2     dt        2

Compute that constant by substituting in the initial conditions then ﬁnd an equation for dx/dt
in terms of x.

Solution. Using part (a), we evaluate the equation at t = 0:
1     1     1
constant = m02 + kA2 = kA2 .
2     2     2
—2—

Next we solve for dx/dt:
2
1      dx        1     1                             dx      k 2
m             = kA2 − kx2               ⇒             =      (A − x2 ).
2      dt        2     2                             dt      m

c) Finding the period. The equation you obtained in part (b) is a differential equation for x(t).
To solve it, we ﬁrst gather terms involving x and dx on the left and terms involving dt on the
right.
Finish solving the differential equation by integrating both sides with respect to correspond-
ing variables. Use your result to determine the period of the oscillator.

Solution. We have
dx
=      dt.
k
m
(A2      − x2 )
Substitute x = A sin u, whereupon dx = A cos u du, giving

m       A cos u du
= t + C,
k       A2 − A2 sin2 u

where C is an arbitrary constant. Next we use the trig identity 1 − sin2 u = cos2 u to get:

m       A cos u du           m                 m
√           =                 du =       u = t + C.
k         A2 cos2 u          k                 k

But u = arcsin(x/A), so that

k
arcsin(x/A) =              (t + C),
m
or
k
x = A sin              (t + C) .
m
Since sine has period 2π, x will go through one cycle when the argument of sine goes through
2π, giving the following expression for the period T :
2π
T =               .
k
m

This period corresponds to a radian frequency of ω =              k/m.

If you want to learn more about problems of this kind, consider taking the following engineering
course: TAM 2030 Dynamics (also ENGRD 2030).

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 44 posted: 1/13/2010 language: English pages: 2