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Unformatted text preview: LINEAR HARMONIC MECHANICAL OSCILLATORS
(Translation) Deﬁnition: A mass M will perform linear harmonic (or simple harmonic) oscillations about a
point of equilibrium (F = 0) if it is acted upon by a force which is proportional to its
__) displacement (x) from equilibrium and always acts in a direction opposite to the displacement
vector; So essential ingredients are:
Magnitude of F proportional to (x)
Direction of F opposite to ( x)
_> Hence F is called: RESTORING FORCE
—) . Alternate Deﬁnition: The object has a potential energy which varies as the square of the
displacement. L.H. oscillations are ubiquitous in the physical world but to start with we consider only 4
realizations of the above deﬁnition. 1. SpringMass Oscillator (Horizontal) war"? A mass M is attached to a spring of constant k. The other end of the spring is attached to a rigid post. M is placed on a smooth frictionless horizontal table such that when it is at x=0, the spring is relaxed so M is in
E m as its weight is supported by N R . If we displace M by an amount x, immediately the spring force
F = —kxfc _) (1) comes into play. So if after displacing M by an amount A we let go, M will be under the
inﬂuence of the force of Eq (1) and will move back and forth as a L.H. oscillator. Why? When you let go, the mass experiences a force that brings it back to 0. But When it
reaches 0 it has a kinetic energy and it cannot stop. So it keeps going until it gets to —A. But new
again it has a force that wants to bring it back to 0. And there you have it. Every time it returns to
E m it fails to stop and when it stops it is not in E m. The corresponding potential energy is 1 2
13:51“ That is: So at x=A, PS}, = 5ch2 . Let go. P wants to reduce but when M reaches 0 (x = O) P is converted 1
[CA2 a ain.
2 g to kinetic energy (éMVmaxz) and that takes it to —A where PW , ‘ 4, , :7 f] Kinetic energy: Energy conservation requires ' ii '. 477:7 ' it, 7 7‘” Z 7.71 Ii: 7 *
lkAZ =lMVzmax =l/Cx2 +‘1‘MV2 2 2 2 2 where Vis the speed when M is at any point x between 0 and A.
So the FORCE equation is F =v/cx)? _) and hence acceleration is
k A
El) ~ —pﬁx ‘ » (3) [Never leave out the minus sign] {it As we showed in class by reference to the case of uniform circular motion, the position,
velocity (magnitude) and acceleration are given by x =Acos(at + 6b) (4/) (73%)
V=—Aa)Sin(a)t+00) ‘ . (5)
az—AmzC0s(a)t+60)=—co2x (6) Where A = amplitude, the largest value of x and is determined by the potential energy stored in
the spring mass system to initiate the motion.
and a) = Angular frequency k
a):—
M and is related to the two fundamental properties of the oscillator. Frequency f = :0— = # of oscillations per second 27: . 1 . . .
Perlod T = — = time taken for one complete oscﬂlation 6 0 = phase and tells you precisely the position of M at t = 0. For example, if 60:0, M is at A at t=0. If 60:”2,Mis atht t=0. WW "" 2. SpringMass Oscillator (Vertical) In this case, of course, as soon as you attach
the mass, the spring will stretch. If you hold
the mass and allow the stretching to happen while you keep holding it until the spring is fully stretched due to the force
— Mg )0 and then let go, the mass will be in E m because spring force lip = —kyy
__)
Will be balancing —— Mg )9
(~10)  Mg) )7 = 0
i.e.
A), = “M; k Ftp ll +9 _> to give Now, if you want M to oscillate you must pull it by an amount A, store l M2 energy in it and it 2 Will oscillate around the newE m pt. Ay = :—:—{g , A above and —A below it. Note that —Mg )9 is a constant force and is no; a restoring force, so it does not affect the period
of the oscillator. All it does is to move the point of E m. k a) is still equal to =
M T=27r\/E
k Alternate expt.
y=0. spring unstretched
Attach M.
Let go.
The mass will drop.
Come to rest.
Rise.
Oscillate. How far is the initial drop?
Determined by conservation of energy. 1 2
lap =5” l
[Pg =ngl 'Aty=0 Pm=0 Pg=0 So now y max é‘kyzmax +Mg ymax = 0 Of course,_=. m pt. is still y = —$
So mass will oscillate between c
y=0and y: “Zilg And again Mg cannot be a restoring force. Only F can do that, so
N) Sp 3. SIMPLE PENDULUM 5? Mass hung from a rigid support using a light string of M
length I [if mass has size 1 must be measured to its C.G.] g
When string is vertical (nearly) mass in E m T— Mg = 0
If you pull mass sideways by amount + 6max and let go it will oscillate between
the 6max and — 6 max . Consider the forces at some angle 6. We can break
~ Mg)? into its components: 1. along the string (radius of circle on which M
travels)
Mg Cos 6
ii. Perpendicular to string (along tangent to circle on which M travels)
FT = —M g Sm 6 2'
__) i“? Because of —ive sign it is certainly a restoring force, so that is good. But it is not
proportional to displacement as it stands. However, if we play our cards right, we can make it so. Recall that when 6 << 1, Sin 6 m 6 So make sure 6 is small! Then P; : ._) And that will give us LH oscillations, Notice: 0 =% Where 3 is displacement along arc. t3 So —M SA
7: g T
_> Z _ SA
a: g r
e l
=—W2sf And therefore angllar frequency must be a) = J; And period becomes T = 27r\/Z
8
As you checked in your experiment. By analogy with case 1 above, we can write
S=Acos(wt+ a),A =16’max Or 9 = 6maxcos(cut+ a)
K: —Amaae cosin(cot+ a) a,=—w2A ‘ Also, potential energy (assuming P = 0, when 6 = O) is
P = Mg 1 (1—0036)
But for 6<<1 2
C056 =l—6—
2
2
So, P= Mgzm Which is what we need for LH oscillators 4. SQhere with a Diametric Hole Take a uniform sphere made of a material of density d.
Make a diametric hole in it (very narrow hole). In the
hole, place a mass m at a distance r from the center of
the sphere. What is the gravitational force on m: FG :—4—ﬂderf a 3
It is a restoring force. It is directly proportional to r. If you let m go it will have LH oscillations with angular frequency «ta: 1/4?” Gd So there! LINEAR HARMONIC OSCILLATORS
(ROTATIONAL) Deﬁnition: To get L.H. Oscillations involving rotation we need 53a torque which is
proportional to the angular displacement (6) from equilibrium 5in is opposite to the
displacement vector. That is, r = —c 6
—> —>
In our case all rotations are about zaxis so
’5 = —c62
_,
And since
I a = 2'
—> —>
C A
a = — — 6 z
> I which by analogy with the prior discussion gives
a = —w262
.__) And so angular frequency Two examples follow, Note that Potential Energy is:
P =
P0 = 1C62 6 V
2 6 = O L SIMPLE PENDULUM Single mass point M moving on a circle of radius 1. At angle 6. rzer
~> _) »)
=—Mngin62 [for case shown] and again we need 6 «1 so that Here I = M12 Soa=§92
—> z
co=‘/E and T=27r\/Z
l g 2; PHYSICAL PENDULUMS and therefore as before. Bar suspended from a rigid support at P. C. G is 1 meters away from P. r=—Mngim92
_, and for 6 «1 r =~Mg162
s,
=10:
_,
so
a —— —Mg162
_) I
yielding
(0: £151 ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 122 taught by Professor Bhagat during the Fall '11 term at Maryland.
 Fall '11
 Bhagat
 Physics

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