Asst_34_Query

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Physics I

05-05-2008

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22:47:35

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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RESPONSE -->

Because if x is too large then the motion will not be simple harmonic motion.

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22:48:10

** 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. **

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23:00:52

What does simple harmonic motion have to do with a linear restoring force of the form F = - k x?

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RESPONSE -->

Simple harmonic motion results from the resistance of the object to change (k for a pendulum or spring for example) and the force that keeps it oscillating outside of its equilibrium point.

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23:01:05

** 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.

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

The only 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), showing that SHM can be modeled by a point moving around a reference circle of appropriate radius C, starting at position `phi when t = 0. **

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23:01:38

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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RESPONSE -->

omega = sqrt (k/m)

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23:01:47

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.

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23:03:20

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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RESPONSE -->

sin(theta)

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23:06:41

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)

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RESPONSE -->

I mixed up the x and y values in my mind... A theta value of 0 needs to equal the full radius, hence cos not sin.

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23:09:20

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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RESPONSE -->

If we know the force constant and the position we can find the total energy by simplifying .5kx^2.

velocity at equilibrium would be found by solving for A * cos(theta)*omega

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14:35:37

** 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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I can tell that I am going to continue to have trouble with pendulums for a little while. I can see all of the same patterns, but they aren't really making sense for me just yet.

Potential energy at position x is .5 k x^2. At extreme position potential energy is .5 k A^2, and since the pendulum is for an instant at rest at that position, the KE is 0. Thus the total energy at this position is .5 k A^2.

Energy is conserved for an ideal pendulum, so at any point we have PE + KE = .5 k A^2, so that .5 k x^2 + KE = .5 k A^2.

Thus KE = .5 k A^2 - .5 k x^2.

At equilibrium x = 0 so KE = .5 k A^2, so 1/2 m v^2 = .5 k A^2. Solve for v.

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Simple overview:

If F = - k x then total energy = .5 k A^2 so PE + KE = .5 k A^2 so .5 mv^2 + .5 k x^2 = .5 k A^2.

If F = - k x then the object's motion is a projection of the motion of a point moving about the unit circle of radius A with angular velocity omega = sqrt(k/m). Supposing that you use the x projection we have

x(t) = A sin(omega * t)

v(t) = -omega * A cos(omega * t)

a(t) = -omega^2 * A sin(omega * t).

These two conditions are equivalent in that if you substitute the above expressions for x and v into the energy equation .5 mv^2 + .5 k x^2 = .5 k A^2 you get an identity (recall that omega = sqrt(k / m), and the identity is not difficult to prove).

There's more to it than this, but this is your context.