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, where k is a positive constant.
Newton’s Second Law:
F = m a

equation of motion

- k x = m a, or a = - (k/m) x d
2 dt
2 x ( t )
 k m x ( t )

0 , or d
2 dt
2 x ( t )
 
2 x ( t )

0 , with

2  k m
¶ 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.
O x m
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, with

2 = k/m.
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P1X — Dynamics Section 6 SHM

(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
O x m
 m mg
Vertical:
Horizontal:
Now
So from (1), and then, from (2),
T cos

- mg = may = 0, if approx horizontal
-T sin

= max sin

= x/L, cos

=

(1- sin 2 
) =

[1- (x/L) 2 ]

1, if x

L , so that

is small.
(1)
(2)
T = mg, ax = - (T/m) sin

= - (g/L) x = -

2 x, 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 = v
0 t + ½ at 2 .
What function x(t) satisfies a = -
 2 x or d 2 dt
2 x ( t )
 
2 x ( t )

0 ?
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 dt
 
A
 sin(

t
 
),
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P1X — Dynamics Section 6 SHM

and acceleration dt
2
 
A
 2 cos(

t
 
)
   2 x
Two special cases are and


= 0,
= -

/2 , x = A cos x = A sin

 t t, which has x =A at t =0,
, which has x = 0 at t = 0 .
x = A cos
 t v = - A

sin
 t a = -

2 x

x oscillates between +A and -A. A is called the amplitude of the SHM.

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.
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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, and acceleration a = -

2 A, radially inwards.
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.
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P1X — Dynamics Section 6 SHM

x = A cos

, v x
= -

A sin

, where a x
= -

2 A cos

= -

2 x

=
 t.
(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 v 2 = (v
0
) 2 + 2ax in the case of constant acceleration a), by using cos 2 
+ sin 2 
= 1,
 2 x 2 + v 2 = A 2  2 [ cos 2 
+ sin 2 
] = A 2  2
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P1X — Dynamics Section 6 SHM

v =
  
(A 2 - x 2 ), 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
Now, mg = kx (equilibrium situation),

x = mg/k.

2 = k/m, and from above,

= 2
 f, = 4

s -1 ,

2 = 16

2 s -2

x = mg/k = g/

2 = 9.8 ms -2 x 0.0063 s 2 = 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 =
 
(A 2 - x 2 ), v max
=

A = 880

Hz x 5 x 10 -4 m = 1.38 ms -1 a = -

2 x, so a max =

2 A = (880

Hz) 2 x 5 x 10 -4 m = 3821.5 ms -2 .
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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

(

L )
 m dt
2
.
The equilibrium position is where the acceleration = 0, which gives x = L + mg/k, = x
0
say.
Let X = x
– x
0
, the displacement from equilibrium, then
 kX
 m dt
2
, so we have SHM about the stretched position.
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P1X — Dynamics Section 6 SHM