Assignment34_OpenQA

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course Phy 241

2/19/2012 1110pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

034. Simple Harmonic Motion

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Question: `q001. Note that this assignment contains 8 questions.

An early experiment in this course demonstrated that the net force restoring a pendulum to its equilibrium position was directly proportional to its displacement

from equilibrium. This was expressed in the form F = - k * x, where x stands for the displacement from equilibrium and k is a constant number called the Restoring

Force Constant, or sometimes a bit more carelessly just the Force Constant.

A current experiment demonstrates that the motion of a pendulum can be synchronized with the horizontal component of a point moving around a circle. If the pendulum

mass is m and the force constant is k, it follows that the angular velocity of the point moving around the circle is `omega = `sqrt( k / m ).

If a pendulum has force constant k = 36 Newtons / meter and mass 4 kg, what is `omega? How long does it therefore take the pendulum to complete a cycle of its

motion?

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Your solution:

omega = sqrt( k / m)

= sqrt(9) = 3 rad/s

T = 2 pi/ 3 rad/s

= 2.09 s

confidence rating #$&*:

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Given Solution:

Since `omega = `sqrt( k / m), we have

`omega = `sqrt( (36 N/m) / (4 kg) ) = `sqrt( 9 (N/m) / kg ) = `sqrt( 9 [ (kg m/s^2) / m ] / kg ) = `sqrt(9 s^-2) = 3 rad/s.

Always remember that this quantity stands for the angular velocity of the point on the reference circle.

[ There is a good reason why we get the radian unit here, but to understand that reason requires a very good understanding of calculus so we're not going to discuss it at this point.]

A cycle of pendulum motion corresponds to a complete trip around the circumference of the circle, an angular displacement of ` pi radians. So if the reference

point is moving around the circle at 3 rad/s, to complete a cycle of 2 `pi rad requires time

T = 2 `pi rad / (3 rad/s) = 2 `pi / 3 sec, or approximately 2.09 sec.

This time is called the Period of Motion of the pendulum, and is customarily designated T.

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Self-critique (if necessary): OK

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Self-critique rating:

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Question: `q002. Recall that a pendulum with mass m and length L experiences a restoring force F = - m g / L * x, so that we have F = - k x with k = m g / L.

What is the period of motion of a pendulum of length 3 meters and mass 10 kg?

What would be the period of a pendulum of length 3 meters and mass 4 kg?

Does your result suggest a conjecture?

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Your solution:

k = m g / L

omega = sqrt( k / m )

omega = sqrt( g / L )

T = 2pi / omega

T = 2pi / sqrt( g / L )

T = 2pi / sqrt( 9.81 m/s^2 / 3 m )

= 3.47 rad/s

mass is irrelevant because is is factored out in along the way

confidence rating #$&*:

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Given Solution:

For a pendulum 3 meters long with mass 10 kg, we have k = m g / L = 10 kg * 9.8 m/s^2 / (3 meters) = 32.7 ( kg m/s^2 ) / m = 32.7 N / m.

The angular velocity of the reference point for this pendulum is thus `omega = `sqrt( k / m ) = `sqrt ( 32.7 N/m / (10 kg) ) = `sqrt( 3.27 s^-2) = 1.81 rad/s.

For a pendulum 3 meters long with mass 4 kg we have k = m g / L = 4 kg * 9.8 m/s^2 / (3 meters) = 13.1 N / m, so `omega = `sqrt( 13.1 N/m / (4 kg) ) = `sqrt( 3.28 s^-2) = 1.81 rad/s.

These angular frequencies appear to be the same; the only difference can be attributed to roundoff errors.

This common angular frequency implies a period T = 2 `pi / `omega = 2 `pi / ( 1.81 rad/s ) = 3.4 sec, approx..

Noting that both pendulums have length 3 meters we therefore conjecture that any pendulum of length 3 meters will have an angular frequency of 1.81 radians/second and period approximately 3.4 sec. We might even conjecture that the period of a pendulum depends only on its length and not on its mass.

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Self-critique (if necessary): OK

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Self-critique rating:

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Question: `q003. What is a symbolic expression for the period of a pendulum of length L and mass m? Hint: Follow the same reasoning steps as in the preceding

example, but instead of numbers use symbols at each step.

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Your solution:

k = m g / L

omega = sqrt( k / m )

omega = sqrt( g / L )

T = 2pi / omega

T = 2pi / sqrt( g / L )

confidence rating #$&*:

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Given Solution:

The reasoning process went like this: We found the restoring force constant k from the length and the mass, obtaining k = m g / L. Then we found the angular frequency `omega = `sqrt( k / m ) using the value we obtained for k. Our result here is therefore `omega = `sqrt( k / m ) = `sqrt( [ m g / L ] / m ) = `sqrt( g / L ).

We note that the mass divides out of the expression so that the angular frequency is independent of the mass. The period is T = 2 `pi / `omega = 2 `pi / (`sqrt ( g / L ) ) = 2 `pi `sqrt( L / g ).

[ If you don't see what's going on in the last step, here are the details: 2 `pi / `sqrt( g / L ) = 2 `pi / [ `sqrt(g) / `sqrt(L) ] = 2 `pi * `sqrt(L) / `sqrt(g) = 2 `pi `sqrt( L / g ) ].

Our expression for the period is also independent of the mass. This would confirm our conjecture that the period of a pendulum depends only on the length of the pendulum and is independent of its mass.

NEARLY CORRECT STUDENT SOLUTION:

k = mg/L

omega = sqrt(k/m)

period of pendulum = 2pi rad / omega

INSTRUCTOR RESPONSE:

Your expressions are correct, but your answer isn't in terms of L and m.

Using your expressions, it's easy to get a result in terms of L and m.

k = m g / L so

omega = sqrt(k / m) = sqrt( (m g / L) / m) = sqrt( g / L).

So

period of pendulum = 2pi rad / omega = 2 pi / (sqrt(g / L ).

This answer would be acceptable for a Physics 121 student, but the denominator involves a fraction, so it isn't quite in standard form.

2 pi / (sqrt(g / L ) = 2 pi * sqrt( L / g), which is standard form.

STUDENT QUESTION: I knew you had to obtain omega somehow. I got a little confused on how omega was equal to sqrt(g/L). Could you explain that a little more

INSTRUCTOR RESPONSE: The net force on a pendulum displaced x units from equilibrium, where | x | < < L (magnitude of displacement is much less than pendulum length), is F_net = - (m g / L) * x.

This was found in previous assignment, using the fact that the vertical component of pendulum tension is very close to m g, and is in the same proportion to the horizontal component as the displacement to the length of the pendulum.

Thus F_net = - k x, where k = m g / L.

Since omega = sqrt(k / m), we have omega = sqrt( (m g / L) / m) = sqrt(g / L).

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Self-critique (if necessary): OK

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Self-critique rating:

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Question: `q004. The frequency of a pendulum is the number of cycles completed per unit of time. The usual unit of time is the second, so the frequency would be

the number of cycles per second. What is the frequency of a pendulum of length 20 cm?

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Your solution:

k = m g / L

omega = sqrt( k / m )

omega = sqrt( g / L )

T = 2pi / omega

T = 2pi / sqrt( g / L )

= 2pi / sqrt(980 cm/s^2 / 20 cm)

= 0.897 rad/s

confidence rating #$&*:

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Given Solution:

We know that the period of a pendulum is T = 2 `pi `sqrt( L / g ). Using L = 20 cm we must use g = 980 cm/s^2 in order to have compatible units in our calculation, we obtain T = 2 `pi `sqrt( L / g ) = 2 `pi `sqrt( 20 cm / (980 m/s^2) ) = 2 `pi `sqrt( .02 s^-2) = 2 `pi * .14 rad/sec = .88 sec (approx).

The period represents the number of seconds required for the pendulum to complete a cycle. To obtain the frequency, which is the number of cycles per second, we take the reciprocal of the period:

f = 1 / T = 1 / (.88 sec / cycle) = 1.14 cycles / sec.

This pendulum will go through 1.14 complete cycles in a second.

STUDENT COMMENT: i didn't know where to start here, i see how you got the answer for T, but had no idea to get frequency by dividing 1 by T

INSTRUCTOR RESPONSE: Be sure to make a note of this. It's not difficult to remember, but if you think in terms of the meanings of the quantities it's easier than trying to keep a bunch of similar-looking formulas straight.

Recall the definitions of frequency and period from the very first experiment of the course, the Introductory Pendulum Experiment. It was not an accident that we started the course with the very last, and perhaps most challenging topic. We have encountered the pendulum at many points in the course, and these last few assignments are designed to complete our understanding of this common system.

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Self-critique (if necessary):

Thought you were asking for period

To obtain the frequency, which is the number of cycles per second, we take the reciprocal of the period:

f = 1 / T = 1 / (.88 sec / cycle) = 1.14 cycles / sec.

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Self-critique rating: 3

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Question: `q005. Early in the course the period of a pendulum was said to be related to its length by the equation T = .20 `sqrt(L), where T is in seconds when L

is in cm. If we rearrange the equation T = 2 `pi `sqrt( L / g ) to the form T = [ 2 `pi / `sqrt(g) ] * `sqrt(L) and express g as 980 cm/s^2, we can simplify the

factor in brackets. Do so and explain how your result confirms the equation given earlier in the course.

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Your solution:

T = [ 2 `pi / `sqrt(g) ] * `sqrt(L)

T = 2pi / sqrt(980 cm/s^2)] * sqrt(L)

T = 0.20 s/cm * sqrt(L)

confidence rating #$&*:

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Given Solution:

The factor in brackets is [ 2 `pi / `sqrt(g) ], which becomes 2 `pi / `sqrt(980 cm/s^2) = 2 `pi / ( 31.3 `sqrt(cm) / s ) = .20 s / `sqrt(cm).

The equation is therefore T = .20 s / `sqrt(cm) * `sqrt(L).

If L is given in cm then `sqrt(L) will be in `sqrt(cm) and the units of the calculation will be seconds.

STUDENT COMMENT:

this factoring and making equations confuse me, it is hard enough to figure out the number to go into given equations

INSTRUCTOR RESPONSE:

Such confusion is typical among Principles of Physics students, is not unusual among General College Physics and even occurs among students in University Physics.

When the expressions get complicated, be sure you write them out on paper. You can't hope to read or operate on a challenging algebra expression without writing it down in standard form.

The algebra involved here approaches but does not exceed the level of a high school Algebra II class and should be