# Simple Harmonic Motion General Formula

• Haths
In summary: Q is greater than the critical angle then the rod will generate over damped motion, if Q is less than the critical angle then the rod will generate undamped motion. End of summary.
Haths

## Homework Statement

Well it's actually a colection of questions that I'd like to know if I've got the right end of the stick in answering;

Data to work with (just making it up):

Period (T)
Mass (m)
Gravitational Accelaration (g) 9.81 ms-2

1) Say we have a disk suspended from a thin light wire lying in the horizontal plane, what is the torsion constant?

2) Say the thin light wire is replaced by an ideal spring, ignore air resistance. What would the spring constant be?

3) Say we know include air resistance modeled with Fj = -bvj where b is ? kg/s for what range of values will the motion be overdamped?

4) What is the quality factor for this osscilation, what does the quality factor tell you?

5) Say we replace the disk with a point mass of the same properties on a light inextensible rod, how long does the rod need to be to generate the same period as in 1) and 2)?

2. The attempt at a solution

1) Say we have a disk suspended from a thin light wire lying in the horizontal plane, what is the torsion constant?

Torque = T
Angular Accelaration = a
Moment of Inertia = I
Angle = Q
1 / T = f

My line of resoning has gone, to make it ossiclate the disk has been rotated from it's equlibrum position by some angle Q and it is experiencing a torque T trying to restore it. Hence by comparing that to Hookes law F=-kx I can write an analogous formula;

T=-kQ

As torque is related to the moment of intera and hence accelaration of the disk back towards its equlibrium possition;

T=Ia
-kQ=Ia

-kQ / I =a

Now in general S.H.M. we have an accelaration that looks like;

[d2x/dt2] = -kx/m

and I have;

[d2Q/dt2] = -kQ/I

Which looks the same. Hence my 'omega factor' frequency constant should be sqrt{ K / I } Thus to find the torsion constant k all I have to do is;

1 / T = (1 / 2PI) * sqrt{ K / I }

Therefore I have;

I*(2PI f)2 = k

Yes?

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

2) Say the thin light wire is replaced by an ideal spring, ignore air resistance. What would the spring constant be?

Well I don't need to be analogous with this one as the spring constant is just;

m*(2PI f)2 = k

If I got the above right...

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

3) Say we know include air resistance modeled with Fj = -bvj where b is ? kg/s for what range of values will the motion be overdamped?

Here I have thought;

ma = -kx - bvj

Therefore;

a = -kx/m - bvj/m

Now I know that the general solution is an exponetial decay with ossilation like;

x = Ae(-b/2m)t * cos( Wt )

Now the b value is going to affect the period as it will slow it down and I looked this up in a textbook and it said that;

w = sqrt{ -b2 / 4m2 + k/m }

But had no explanation why this was (I guess that's because it's part of the solution to the differentail equation I have above, but it doesn't go through the proof in that book)

Ok so what is overdamped well I'm preaty sure it's when there are no ossiclations in the system, and critically damped is when it returns to it's equlibrium position in the smallest time possible, yes?

Hence I would need a W that is equal to 0, because that would get rid of that whole section. So;

b2 / 4m2 = k / m
b2 = sqrt{ 4km }
b = sqrt{ 2 sqrt{ km}}

So b is greater than sqrt{ 2 sqrt{ km}}, yes?

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

4) What is the quality factor for this osscilation, what does the quality factor tell you?

The quality factor I believe is given by;

G = W' / (b/m)

Which I guess means;

G = sqrt{ k/m } / ( sqrt{ 2 sqrt{ km}} / m )

But I havn't a clue really because we haven't covered this yet, so I can't even answer the second part, although I would guess it is a ratio of the frequencies and would tell you somthing about how many osscilations will occur in the period of the time constant.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

5) Say we replace the disk with a point mass of the same properties on a light inextensible rod, how long does the rod need to be to generate the same period as in 1) and 2)?

If we displace the mass by some angle Q, then the transversail force will be mg sin(Q). Thus the torwue will be;

T = rmg sin(Q)

Now the only thing I could think of at this point is to draw a right angled triangle between the point mass, the point it is swinging from and the equilibrum possition. Doing so we can represent the particles ossiclation with a new quanta say x from the equilibrium line. Thus;

T = rmg sin(Q)

sin(Q) = x/r

. => T = rmg x/r
. = mg x

Going all the way back to my first question let the torque = Ia, then divide by the moment of inertia of this system which will just be mr2. Therefore;

[d2Q/dt2] = gx / r2

Which like question 1) if you can use that as an analogous form, can be rerranged to;

r = sqrt{ g / (2PI f)2 }

To find the length of the pendulum, yes?

Cheers,
Haths

NOTE: If you want me to write down the steps of the rearrangements I can.

Hi,

I've had time to look at #'s 1, 2, and 3. 1 and 2 look right. For 3, see my comment below:

Haths said:
3) Say we know include air resistance modeled with Fj = -bvj where b is ? kg/s for what range of values will the motion be overdamped?

Here I have thought;

ma = -kx - bvj

Therefore;

a = -kx/m - bvj/m

Now I know that the general solution is an exponetial decay with ossilation like;

x = Ae(-b/2m)t * cos( Wt )

Now the b value is going to affect the period as it will slow it down and I looked this up in a textbook and it said that;

w = sqrt{ -b2 / 4m2 + k/m }

But had no explanation why this was (I guess that's because it's part of the solution to the differentail equation I have above, but it doesn't go through the proof in that book)

Ok so what is overdamped well I'm preaty sure it's when there are no ossiclations in the system, and critically damped is when it returns to it's equlibrium position in the smallest time possible, yes?

Hence I would need a W that is equal to 0, because that would get rid of that whole section. So;

b2 / 4m2 = k / m
b2 = sqrt{ 4km }

There's an algebra mistake here, this should be:

b2 = 4 m2 k / m = 4 k m

b = sqrt{ 2 sqrt{ km}}

So b is greater than sqrt{ 2 sqrt{ km}}, yes?

As Homer Simpson would say; "Doh!" cheers ;)

Haths

## 1. What is the general formula for simple harmonic motion?

The general formula for simple harmonic motion is x(t) = A*cos(ωt + φ), where x(t) is the displacement of the oscillating object, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

## 2. How is the amplitude of a simple harmonic motion related to the maximum displacement?

The amplitude of a simple harmonic motion is equal to the maximum displacement from the equilibrium position. In other words, it represents the distance from the center to either end of the oscillation.

## 3. What does the angular frequency represent in the simple harmonic motion formula?

The angular frequency, ω, represents the rate at which the object oscillates back and forth. It is equal to 2π divided by the period of the motion (T).

## 4. How does changing the phase angle affect the motion in simple harmonic motion?

Changing the phase angle shifts the starting point of the oscillation. It does not affect the amplitude or frequency of the motion, but it changes the position of the object at t=0.

## 5. Can the simple harmonic motion formula be used for all types of oscillations?

No, the simple harmonic motion formula can only be used for objects that experience a restoring force proportional to the displacement from the equilibrium position. This includes pendulums, springs, and certain types of waves.

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