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Text assignments:

General College Physics:  Chapter 11, Sections 1-4.  Do problems 7, 14, 24, 32, 33

University Physics:

Experiment:  Chain between 5 and 10 rubber bands together.

Using the given mass figure out the force constant for your rubber bands.

Predict how far the given mass should stretch your rubber bands from an existing equilibrium position.

Analysis:

Your observations should have given you a value of omega.

Using omega = sqrt(k/m) you sho

From last time:

Where does omega = sqrt(k / m) come from?

If Fnet = - k x, then since Fnet = m a we have

m a = - k x so that

a = - k / m * x.

a and x are the acceleration and position functions. 

• If we had a graph of x vs. t we could find the slopes of this graph and construct a graph of v vs. t.
• If we had a graph of v vs. t we could find the slopes of this graph and construct a graph of a vs. t.

So

• The v function is the derivative with respect to clock time of the x function and
• The a function is the derivative with respect to clock time of the v function so
• The a function is the second derivative with respect to clock time of the x function.

So

• a = x ''

and our equation

• a = - k / m * x

becomes

• x '' = -k / m * x.

Everyone should understand this development, whether you've had calculus or not.  You know that slope-taking is referred to a taking the derivative.  a = x '' just means you do this process twice to the x vs. t graph and end up with the a vs. t graph.

University Physics question:

What sorta functions have second derivatives that are multiples of the functions themselves?  Research this question, be sure you understand all possible answers, be sure you understand the chain rule, and be prepared next time.

For this time:

University Physics students have claimed that

• exponential functions,
• sine functions and
• cosine functions

all have second derivatives that are multiples of themselves.  This is correct.

University Physics Questions

What is the second derivative of e^(omega * t)?

What is the second derivative of sin(omega * t)?

What is the second derivative of cos(omega * t)?

What functions satisfy the equation x '' = - k / m * x?

What is the position of the vertical line through the point whose angular position is theta on a reference circle of radius A?

The position of the vertical line is x = A cos(theta), by the definition of the cosine function.

Note that the y coordinate ain't worth a tinker's toot for locating the position of a vertical line.

What is the angular position of the point on the reference circle of radius A at clock time t, given that angular position is 0 when t = 0 and the angular frequency is omega?

For example if angular frequency is 2 rad / sec, what is angular position after 3 sec?  We immediately get angular position 6 radians.

What did we do to get that?

We multiplied the angular frequency by the time interval.

If we know omega and t, then the time interval since t = 0 is just t so the angular displacement since t = 0 is omega * t.  If we started at angular position theta = 0, then our angular position at clock time t is just

• theta = omega * t.

What is the position of the vertical line at clock time t?

The vertical line will be at position x = A cos(theta), which is just

• x(t) = A cos(omega * t).

What is the velocity of the vertical line at clock time t?

The velocity of the vertical line is the derivative with respect to clock time of its position.  So the velocity of the vertical line is

• v(t) = x ' (t) = - omega * A sin(omega * t).

What is the acceleration of the vertical line at clock time t?

• a(t) = v ' (t) = -omega^2 A cos(omega * t).

It's worth noting again that a(t) = x '' ( t).

If k = 10 N / m and mass is .3 kg then what is the angular frequency of the motion of the mass?

omega = sqrt(k / m) = sqrt( ( 10 N/m ) / (.3 kg) ) = 5.7 rad/s, approx..

Assume an amplitude of .5 meters.  If the motion starts at angular position 0 when t = 0, then what is the angular position at t = .4 sec?

The angular position is theta = omega * t = 5.7 rad/s * .4 sec = 2.28 radians, in the second quadrant.

What is the position of the vertical line when t = .4 sec?

The position is x(t) = A cos( omega * t).  A = .5 m and omega = 5.7 rad/s.  So

• x(.4 sec) = .5 m cos( 5.7 rad/s * .4 s) = .5 m cos( 2.28 rad) = -.28 m, approx.

What is the velocity of the vertical line when t = .4 sec?

The velocity of the vertical line is v(t) = - omega A sin(omega t) so we have

• v(.4 sec) = (- 5.7 rad / s * .5 m) sin( 5.7 rad / s * .4 sec) = ... = -2.3 m/s (?).

What is the acceleration of the vertical line when t = .4 sec?

The acceleration of the vertical line is a(t) = - omega^2 A cos(omega t) so we have

• a(.4 sec) = (- 5.7 rad / s) ^ 2 * .5 m * cos( 5.7 rad / s * .4 sec) = ... = 10 m/s^2 (?).

What was the velocity with which the pearl pendulum struck the meter stick?

What was the angular momentum of the pearl before collision, relative to the axis of rotation?

What was the angular momentum of the meter stick immediately after collision?

What was the angular momentum of the pearl immediately after collision?

Was angular momentum conserved?