Query 34

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course PHY 121

034. `query 34

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Question: `q Query Class Notes #33 Why do we say that a pendulum obeys a linear

restoring force law F = - k x for x small compared to pendulum length?

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Your solution:

The force is directly related to the displacement of the pendulum. The restoring force

is the force that is working toward equilibrium. Equilibrium for the pendulum is

hanging straight down, not moving, with the force of gravity times the mass of the

weight on the end of the pendulum. When you pull the pendulum back and release it, it

accelerates toward the equilibrium position, then slows until the velocity is 0. At

that point it is at the top of the swing, and accelerates downward, back toward

equilibrium. As it hits the equilibrium mark, it has gained sufficient velocity to

swing back up toward the place where it was released.

confidence rating #$&*:

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Given Solution:

`a** The vertical component of the tension in the string is equal to the weight m * g of

the pendulum. At angle `theta from equilibrium we have T cos(`theta) = m * g so T = m *

g / cos(`theta).

The horizontal component of the tension is the restoring force. This component is T sin

(`theta) = m * g * sin(`theta) / cos(`theta) = m * g * tan(`theta).

For small angles tan(`theta) is very close to `theta, assuming `theta to be in radians.

Thus the horizontal component is very close to m * g * `theta.

The displacement of the pendulum is L * sin(`theta), where L is pendulum length. Since

for small angles sin(`theta) is very close to `theta, we have for small displacements x

= displacement = L * `theta.

Thus for small displacements (which implies small angles) we have to very good

approximation:

displacement = x = L `theta so that `theta = x / L, and

restoring force = m * g * `theta = m * g * x / L = ( m*g/L) * x. **

ALTERNATIVE EXPLANATION:

This relationship can be easily visualized in terms of similar triangles. Sketches can

also be found in the appropriate Class Notes.

First:

The angle of the pendulum with vertical is small. It follows that the vertical

component of the tension force is very nearly the same as the tension force.

The vertical component of the tension force is what prevents the mass at the end of the

pendulum from falling. That mass doesn't move significantly in the y direction, so it

is in vertical equilibrium. We conclude that the vertical component of the tension has

magnitude m g, and that the tension therefore has magnitude m g.

Now consider two triangles (which unless you have excellent powers of visualization you

should sketch as you read the description):

The first triangle has as its hypotenuse the length of the pendulum, and its legs are

horizontal and vertical.

The second triangle has as its hypotenuse the tension force acting on the pendulum mass,

and its legs are horizontal and vertical.

The tension force acts in the direction of the string, so the string makes the same

angle with horizontal as the tension vector. We conclude that the triangles are

similar.

The horizontal leg of the first triangle corresponds to the displacement of the pendulum

in the horizontal direction. Call this leg x.

The horizontal leg of the second triangle corresponds to the component of the tension

force in the horizontal direction. Call this leg T_horiz.

So the first triangle has horizontal leg x and hypotenuse L, where L is the length of

the pendulum.

The second triangle has horizontal leg T_horiz and hypotenuse T = m g.

Since the triangles are similar, we have the proportion

x / L = T_horiz / mg

which we solve easily to find that

T_horiz = mg * (x / L) = (m g / L) * x.

The horizontal component T_horiz of the tension is in the direction opposite the

horizontal displacement x.

The horizontal component of the tension is the force that restores the pendulum to

equilibrium.

Thus

restoring force = - (m g / L) * x, or

F_restoring = - k x, with k = m g / L

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Self-critique (if necessary):

The alternative explanation makes more sense to me--the angle of the pendulum is small

which casues the vertical force to be almost the same as the tension force. You can

picture this with a right triangle that has a small angle, which would represent the

displacment angle (theta). The constant in the force of the pendulum is mass times

gravity divided by length of the pendulum. This would represent equilibrium. The

amount of force is proportional to the size of the displacement.

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Self-critique Rating: 2

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Question: `q What does simple harmonic motion have to do with a linear restoring force

of the form F = - k x?

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Your solution:

A pendulum will hang straight down (Force = mass * gravity) at a point of equilibrium

unless there is a displacement. The pendulum swings back and forth through its point of

equilibrium. The outside limits of the pendulum's swing is the displacment. I think of

the simple harmonic motion as swinging, or in the case of a spring bouncing, back and

forth, trying to achieve equilibrium, but unable to because of momentum that is gathered

in the back and forth motion.

confidence rating #$&*:

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Given Solution:

`a** the essential relationship is F = - k x; doesn't matter if it's a pendulum, a mass

on an ideal spring, or any other system where net force is a negative constant multiple

of the displacement from equilibrium.

For Principles of Physics and General College Physics students the following statement

summarizes the relationship:

If a constant mass m is subjected to a net force F = - k x, then if that mass is in

equilibrium at x = 0 it will remain there.

Otherwise it will undergo simple harmonic motion with angular frequency omega = sqrt( k

/ m ).

The following characteristics of the motion of the oscillator are not expected at this

stage, but they are worth noting and thinking about:

Its total energy will be 1/2 k A^2, where A is the amplitude of its motion (its maximum

displacement from the x = 0 position).

As it passes its equilibrium position all its energy will all be kinetic.

At its maximum displacement from equilibrium its energy will all be potential.

Its potential energy at position x is 1/2 k x^2. The rest of its total energy is in the

form of kinetic energy.

University Physic students should also understand the reasons for this:

x is regarded as a function of clock time t, and the expression x '' indicates the

second derivative of x with respect to clock time t.

F = m * a = m * x'', so F = - k x means that

m * x'' = - k x.

The only real-valued functions whose second derivatives are constant negative multiples

of the functions themselves are the sine and cosine functions.

We conclude that x = A sin(`omega t) + B cos(`omega t), where `omega = `sqrt(k/m).

For appropriate C and `phi, easily found by trigonometric identities, A sin(`omega t) +

B cos(`omega t) = C sin(`omega t + `phi).

Thus SHM can be modeled by a point moving around a reference circle of appropriate

radius C, starting at position `phi when t = 0. **

STUDENT QUESTION

My question is, how does this answer the question? Does omega determin

F somehow?

INSTRUCTOR RESPONSE

The brief answer to the question is that when F_net = - k x, the object will undergo

simple harmonic motion, and vice versa (i.e., if the object undergoes SHM, then F_net =

- k x).

The given solution goes into more detail about why this is so.

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Self-critique (if necessary): OK

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Self-critique Rating: 2

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Question: For a simple harmonic oscillator of mass m subject to linear net restoring

force F = - kx, what is the angular velocity `omega of the point on the reference

circle?

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Your solution:

You can find `omega by taking the square root of k/m.

confidence rating #$&*:

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Given Solution:

STUDENT RESPONSE: omega= sqrt (k/m)

INSTRUCTOR COMMENT: Good. Remember that this is a very frequently used relationship for

SHM, appearing in one way on another in a wide variety of problems.

STUDENT COMMENT:

i got this one right, but i thought you wanted it in m/s

INSTRUCTOR RESPONSE: omega is the angular velocity, which is in rad / sec.

If you want the speed of the point on the reference circle, then you would multiply the

angular velocity by the radius of the reference circle. So for example if the radius is

in meters, the speed of the point would be in meters / second.

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Self-critique (if necessary): OK

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Self-critique Rating: 2

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Question: `q If the angular position of the point on the reference circle is given at

clock time t by `theta = `omega * t, then what is the x coordinate of that point at

clock time t?

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Your solution:

This is one where you have to find the cosine because cosine means x. I know that you

put radius cosine ________. But I always get mixed up on what fills in the blank.

confidence rating #$&*:

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Given Solution:

Since theta=omega t, if we know t we find that x = radius * cos(theta) or more

specifically in terms of t

x = radius*cos(omega*t)

STUDENT QUESTION

I understand that theta = omega*t was derived from the concept of velocity = distance

/time. But I don't understand how

they are finding the x coordinate.

INSTRUCTOR RESPONSE

A vector whose length is equal to the radius, which makes angle theta with the positive

x axis, has x and y components radius * cos(theta) and radius * sin(theta),

respectively.

Thus x = radius * cos(omega * t) and y = radius * sin(omega * t).

You should of course sketch all this out on paper. An unlabeled sketch depicting a

typical vector and its x and y components is depicted in the figure below:

STUDENT QUESTION

To find the x coordinate on the circle, we rely on theta=omega t to help us out. If we

know t, then we can use x=

radius*cos(theta) which then becomes a=radius*cos(omega*t)

I flipped through a chapter in the book as well and saw an example that related to

positions as a function of time. It had

x=Acos theta. Could this formula apply to this situation as well?

Is it essentially the same formula or is it for something completely different?

And what exactly does the A represent?

INSTRUCTOR RESPONSE

A represents the amplitude of motion, which is equal to the radius of the reference

circle.

It is the same formula.

The radius of the reference circle is equal to the amplitude of the motion, and the

angle is omega * t.

So x = radius * cos(ometa * t) is exactly the same as x = A * cos(theta).

Recall from Intro Prob Set 5 that the x component of a vector of magnitude A and angle

theta is v_x = A cos(theta). So this is nothing new.

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Self-critique (if necessary):

angle

angle

angle

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Self-critique Rating: 2

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Question: `q Query introductory problem set 9, #'s 1-11 If we know the restoring force

constant, how do we find the work required to displace the oscillator from its

equilibrium position to distance x = A from that position? How could we use this work to

determine the velocity of the object at its equilibrium position, provided we know its

mass?

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Your solution:

If I am looking for the angular velocity, I can use the restoring force constant and the mass (`sqrt k/m).

Work = Force * displacement

Kinetic energy = .5 * Inertia * `omega^2.

Beyond that, I don't know what this is asking.....

confidence rating #$&*:

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Given Solution:

`a** You can use the work 1/2 k A^2 and the fact that the force is conservative to

conclude that the max PE of the system is 1/2 k A^2. This PE will have transformed

completely into KE at the equilibrium point so that at the equilibrium point KE = .5 m

v^2 = .5 k A^2. We easily solve for v, obtaining

v = `sqrt(k/m) * A. **

STUDENT COMMENT: I'm a little confused by that 1/2 k A^2.

INSTRUCTOR RESPONSE:

That is the PE at x = A. To directly reason out the expression PE = .5 k A^2 we proceed

as follows:

PE = work done by system in moving from equilibirum * displacement = fAve * `ds.

The force exerted on the system at position x = A is -k A. The force exerted at position

x = 0 is 0. Force changes linearly with position. So the average force exerted on the

system is

( 0 - kA) / 2 = -1/2 k A.

The force exerted by the system is equal and opposite, so

fAve = 1/2 k A.

The displacement from x = 0 to x = A is `ds = A - 0 = A.

We therefore have

PE = fAve `ds = 1/2 k A * A = 1/2 k A^2.

This is also the area beneath the F vs. x graph between x = 0 and x = A. That region is

a triangle with base A and altitude k A so its area is 1/2 base * height = 1/2 k A^2.

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Self-critique (if necessary):

I have no deep understanding of this whole section with circles, pendulums, and springs. I am able to look through notes, find formulas and how I solved prior problems, and then plug things together for partial understanding, but I am not really mastering this. It feels like I'm reading Greek.

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Self-critique Rating: 1

@&

That's the first step.

I have a grandchild due in October who will be half Greek, 1/4 Italian, and 1/4 whatever I am. Hopefully Greek isn't a bad place to start.

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Question: `q Query Add comments on any surprises or insights you experienced as a result

of this assignment.

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#*&!

&#Good work. See my notes and let me know if you have questions. &#