OSCILLATIONS: II

1. In this chapter we shall continue with the concepts developed in the previous chapter
that relate to simple harmonic motion and the simple harmonic oscillator (SHO). It is
really very amazing that the SHO occurs again and again in physics, and in so many
different branches.
2. As an example illustrating the above, consider a mass suspended
a string. From the diagram, you can see that sin . For small
values of we know that sin . Using (length of arc),
F mg
x L
θ
θ θ θ θ
= −
≈ = we
have . So now we have a restoring
force that is proportional to the distance away from the equilibrium
point. Hence we have a SHO with / . What if we had not
F mg mgx mg x
L L
g L
θ
ω
= − = − = −⎛⎜ ⎞⎟
⎝ ⎠
=
made the small approximation? We would still have an oscillator
(i.e. the motion would be self repeating) but the solutions of the
differential equation would be too complicated to discuss
θ
here.
3. If you take a common object (like a piece of cardboard) and pivot
it at some point, it will oscillate when disturbed. But this is not the
simple pendulum discussed above because all the mass is not
concentrated at one point. So now let us use the ideas of torque and
angular momentum discussed earlier for many particle systems.
You can see that τ = −Mgd sinθ. For small θ ,
2
2
2 2
2 2
sin and so
. But we also know that where I is the moment of
inertia and is the angular acceleration, . Hence, we have
, or, . From
Mgd I
d
dt
Id Mgd d Mgd
dt dt I
θ θ
τ θ τ α
θ
α α
θ θ
θ θ
≈
= − =
=
= − = −⎛⎜ ⎞⎟
⎝ ⎠
this we immediately
see that the oscillation frequency is . Of course, we have
used the small angle approximation over here again. Since all variables
except are known, we can u
Mgd
I
I
ω =
se this formula to tell us what is about
any point. Note that we can choose to put the pivot at any point on the
body. However, if you put the pivot exactly at the centre of mass then
i
I
t will not oscillate. Why? Because there is no restoring force and the
torque vanishes at the cm position, as we saw earlier.
m
θ
L
x = Lθ
T
θ
mg sinθ
4. Suppose you were to put the pivot at point P which is at a distance from the centre
of mass of the irregular object above. What should be so that you get the same
formula as for a simple
L
L
pendulum?
Answer: 2 2
P is then called the centre of gyration - when suspended from this point it appears as if
all the mass is concentrated at the cm position.
5. Sum of two sim
T L I L I
g Mgd Md
= π = π ⇒ =
( ) 1 1 2 2
1 2
1 2 1
ple harmonic motions of the same period along the same line:
sin and sin
Let us look at the sum of and ,
s
x A t x A t
x x
x x x A
= ω = ω +φ
= + = ( )
( ) ( )
2
1 2 2
1 2 2
1 2 2
in sin
sin sin cos sin cos
sin cos cos sin
Let cos cos and sin sin . Using some simple trigonometry,
you can put
t A t
A t A t A t
t A A t A
A A R A R
x
ω ω φ
ω ω φ φ ω
ω φ ω φ
φ θ φ θ
+ +
= + +
= + +
+ = =
( )
( )
( )
2 2 2
1 2 1 2
1 2
2 2 2
1 2 1 2 1 2 1 2
1 2
in the form, sin . It is easy to find R and :
cos and tan sin .
cos
Note that if 0 then and tan 0
0. So we get sin .
x R t
R A A AA A
A A
R A A AA A A A A
x A A t
ω θ θ
φ
φ θ
φ
φ θ
θ ω
= +
= + + =+
= = + + = + = + =
⇒ = = +
( )
( )
2 2 2
1 2 1 2 1 2 1 2
1 2
This is an example of
If then and tan 0
0. Now we get sin . This is .
6. Composition of two simple ha
constructive
interference. R A A A A A A A A
x A A t destructive interference
φ π θ
θ ω
= = + − = − = − =
⇒ = = −
( )
2 2
rmonic motions of the same period but now at right
angles to each other:
Suppose sin and sin . These are two independent motions. We
can write sin and cos 1 / .
Fr
x A t y B t
t x t x A
A
ω ω φ
ω ω
= = +
= = −
2 2
2 2
2
2 2
om this, sin cos sin cos cos sin 1 / . Now square and
rearrange terms to find:
2 cos sin
This is the equation for an ell
y t tx x A
B A
x y xy
A B AB
ω φ φ ω φ φ
φ φ
= + = + −
+ − =
ipse (see questions at the end of this section).
7. If two oscillations of different frequencies at right angles are combined, the resulting
motion is more complicated. It is not even periodic unless the two frequencies are in
the ratio of i
( )
ntegers. This resulting curve are called Lissajous figures. Specifically, if
sin and sin , then periodic motion requires integers.
You should look up a book for more details.
x
x y
y
x A t y B t ω
ω ω φ
ω
= = + =
8. : Typically the frictional force due to air resistance, or in a
liquid, is proportional to the speed. So suppose that the damping force (why
negative sign?). No
b dx
dt
= −
Damped harmonic motion
2
2
2
2
w apply Newton's law to a SHO that is damped:
Rearrange slightly to get the equation for a damped SHO: 0.
kx b dx m d x
dt dt
md x bdx kx
dt dt
− − =
+ + =
( )
2
/ 2
2
Its solution for is cos ' . The frequency is now
2
changed: ' . The damping causes the amplitude to decrease with
2
time and when / 2 1, the amplitud
bt m
m
k b x x e t
m m
k b
m m
bt m
ω φ
ω
≥⎛⎜ ⎞⎟ = − +
⎝ ⎠
= −⎛⎜ ⎞⎟
⎝ ⎠
= e is 1/ 1/ 2.7 of its initial value.
9. There is a characteristic value of the driving
frequency at which the amplitude of oscillation is a maximum. This condit
e
ω
≈
Forced oscillation and resonance.
0 0
0
2
2 0
ion is
called resonance. For negligible damping resonance occurs at . Here is
the natural frequency of the system and is given by . The equation of
motion is: cos
k
m
md x kx F
dt
ω ω ω
ω
=
=
+ =
( ) 0
2 2
0
0
. You should check that this is solved by putting
cos (just subsitute into the equation and see!). Note that the
amplitude "blows up" when . This is because we have no dampin
t
x F t
m
ω
ω
ω ω
ω ω
=
−
→
0
g term here.
With damping, the amplitude is large when ω →ω but remains finite.
m
k