mass reaches the center, its speed causes it to continue moving and the
spring begins to compress. The compression of the spring causes a force
to be exerted towards the center and opposite the motion. As before,
the opposing force leads to the mass slowing down, stopping, and then
returning back to the center where the process begins again.
To formally describe the motion, the force exerted on the spring is related
to the displacement of the mass, x, from the equilibrium position:
F=−kx (11.1)
The negative sign is included so that the force is always directed towards
the center. The value of x =0 is defined as the equilibrium position
where the spring is neither stretched nor compressed and there is no force
exerted on the mass. By Newton’s laws, force is equal to the product of
the mass and acceleration, a, of a particle:
F=ma (11.2)
The acceleration of a particle is equal to the change of velocity, 9 , of the
particle with respect to time. Since the velocity is equal to the change in
position with respect to time, we can write:
(11.3)
We can equate the two expressions for the force, eqns 11.1 and 11.2, and
derive the following second-order differential equation that describes the
motion of the mass:
(11.4)
As we saw for the other problems that we have already studied, the solu-
tions to this equation are of the form sin t, cos t, and et. For this case, the
most useful expression is to use the sine function, including the amplitude,
A, and the frequency ω:
x=Asinωt (11.5)
To verify that this is a correct solution, we will put this solution into
eqn 11.3. To do so, we need to first calculate the second-order derivative:
(11.6)
d
d
d
d
d
d
d
d
2
2
x
ttt
At
t
= sin (A cos
⎛
⎝
⎜⎜
⎞
⎠
ωωω⎟⎟=+ ttA)sin=−ωωω^22 t x=−
m
x
t
kx
d
d
2
2 =−
Fmam
t
m
x
t
== =
d
d
d
d
9 2
2
222 PART 2 QUANTUM MECHANICS AND SPECTROSCOPY
