Download 6) Simple Harmonic Motion

¶ 6) Simple Harmonic Motion
Very important - prototype for all oscillations and waves.
¶ 6.1) Definition
SHM is motion, in which a particle is acted on by a force proportional to its displacement from a fixed point, and in the opposite direction to the displacement.
F = - k x,
Newton’s Second Law:
where k is a positive constant.
F=ma

- k x = m a, or a = - (k/m) x
2
k
d
k
d2
 equation of motion
,
or
x
(
t
)

x
(
t
)

0
x(t )   2 x(t )  0 , with  2 
2
2
m
m
dt
dt
¶ 6.2) Examples of SHM
(a)
Mass m vibrating on an elastic spring, with spring constant k. Let the position of
m, when the string is unstretched, be the origin x = 0.
m
O
x
Consider case of horizontal motion on a frictionless table. (The same result holds for
vertical oscillations, but the analysis is slightly trickier, since we have to include gravity
as well as the elastic force.)
F = - k x,

by Hooke's Law
a = F/m = - (k/m) x = - 2 x,
1
P1X — Dynamics Section 6 SHM
with 2 = k/m.
(b)
Simple pendulum: Point mass m on a string of length L. The string has to be long
and the oscillations small enough for the mass to move almost in a horizontal line.
T


L
m
O
m
mg
x
Vertical:
Horizontal:
T cos  - mg = may = 0,
-T sin  = max
Now
sin  = x/L,
cos  =  (1- sin2  ) = [1- (x/L)2 ]  1,
if x  L , so that  is small.
So from (1),
T = mg,
and then, from (2),
ax = - (T/m) sin  = - (g/L) x = - 2 x,
if approx horizontal
with  = (g/L).
¶ 6.3) Solution for x and v
Note that in SHM acceleration a is not constant, so we cannot use v = vo + at,
x = v0 t + ½ at2.
What function x(t) satisfies
a=-
2
x
or
d2
x(t )   2 x(t )  0 ?
2
dt
Calculus method:
Try x = A cos (t + ),
where A and  are constants, because the derivative of cos is
-sin, and the derivative of sin is +cos.
Then, velocity
dx
  A  sin(  t   ),
dt
2
P1X — Dynamics Section 6 SHM
(1)
(2)
d2x
and acceleration dt 2
  A  2 cos(  t   )
  2 x
Two special cases are
and
 = 0,
 = - /2,
x = A cos t, which has x =A at t =0,
x = A sin t, which has x = 0 at t = 0.
x = A cos t
v = - A sin t
a = - 2 x

x oscillates between +A and -A.

Since A cos [t + ] = A cos [(t +2/) + ] = A cos [(t +T) + ], the motion repeats
after a time T, where T = 2/.
T is called the period of the SHM.
A is called the amplitude of the SHM.
3
P1X — Dynamics Section 6 SHM

The number of oscillations per second is called the frequency, f = 1/T, with units
s-1, also called hertz (Hz).

 = 2/T = 2f, is called the angular frequency, and its units are radians/s.

 is called the phase constant of the SHM. Its value depends on where the particle
is in its oscillation at t = 0, the time we start the clock. It is only important when we
try to add two simultaneous SHM's. e.g. two out of phase sound waves.
¶ 6.4) Relation to uniform circular motion
This is important in its own right, but also gives a non-calculus derivation of the above
results.
Consider a mass m at P moving round a circle of radius A with constant angular velocity
. It has tangential speed
v = A,
along the tangent to the circle,
a = - 2 A,
radially inwards.
and acceleration
Now consider the projection of this motion on a diameter.
Sketch:
If the chosen diameter is the X-axis, this is the point N.
As P goes round the circle, N oscillates along the diameter between x= + A and x = - A,
so the radius A is the amplitude of the oscillation. The motions of both P and N have
the same period T. By definition, the angular speed  = angle/time = 2/T, so T = 2/.
We have still to show that the motion of N is SHM, i.e. that a = - 2 x. To do this, resolve the velocity and acceleration vectors of the point P on the circle and look at the xcomponents, since these are the same for P and N.
4
P1X — Dynamics Section 6 SHM
x = A cos ,
where  = t.
vx = -  A sin ,
ax = - 2 A cos  = - 2 x
(q.e.d.)
This shows that N does indeed move with SHM.
Phase constant:
To get the solution x = A cos t, we started our clock so that t = 0 when the point P was
at x = A.
If we had chosen t = 0 when P was at angle  round the circle, then x = A cos(t +).
This is the most general form.
Diagrams:
Note the formulae for the period in the cases of section 6.2:
T = 2/ = 2 (m/k),
for spring,
T = 2 (L/g),
for simple pendulum,
with the rather surprising features:
(a) The period is independent of the amplitude, (which enables a pendulum to be used
as a clock).
(b) For the simple pendulum the mass m does not matter either.
¶ 6.5) Velocity–displacement equation for SHM
Note that we can obtain an equation relating x and v, without involving t (the analogue
of v2 = (v0)2 + 2ax in the case of constant acceleration a), by using cos 2  + sin2  = 1,
2x2 + v2 = A2 2 [ cos2  + sin2  ] = A2 2
5
P1X — Dynamics Section 6 SHM
v =   (A2 - x2),
showing that
v is greatest, =  A, when x = 0, i.e. at the centre of the oscillation,
and
v = 0 at x =  A,
at the extremities.
¶ 6.6) Examples
(a)
10 kg mass hanging on a spring has characteristic frequency 2 Hz. How much
will the length of the spring change when the mass is detached?
Equivalent question: How much is it stretched, when hanging in the equilibrium position?
In this position
mg = kx (equilibrium situation),
Now,
2 = k/m,
and from above,
 = 2f, = 4 s-1,
 x = mg/k.
2 = 162 s-2
 x = mg/k = g/2 = 9.8 ms-2 x 0.0063 s2 = 6.2 cm.
Note that we don't need to know the mass - the answer is the same for all m.
(b)
A point on the end of a 440 Hz tuning fork vibrating with SHM moves a total distance of 1 mm from one extreme position to the other. What is the maximum speed and
maximum acceleration of this point?
We have  = 2f = 880 Hz, and A= 0.5 mm = 5 x 10-4 m.
v =  (A2 - x2),
vmax =  A = 880 Hz x 5 x 10-4 m = 1.38 ms-1
a = - 2 x,
so amax = 2 A = (880 Hz)2 x 5 x 10-4 m = 3821.5 ms-2.
6
P1X — Dynamics Section 6 SHM
¶ 6.7) Appendix: spring oscillating vertically
Take x to be the coordinate of the mass as measured from the point of support.
Let the natural, i.e. unstretched length of the spring = L
O
x
m
The equation of motion is
mg  k ( x  L )  m
d2x
.
dt 2
The equilibrium position is where the acceleration = 0, which gives
x = L + mg/k, = x0 say.
Let X = x – x0, the displacement from equilibrium, then
 kX  m
d2X
dt 2
,
so we have SHM about the stretched position.
7
P1X — Dynamics Section 6 SHM