cq_1_101

Your 'cq_1_10.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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What you did was good, but you didn't use it to answer the question as posed.

See the following commentary.

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The following is a solution to the given problem.

Please compare the given solution with your solution and submit a

self-critique of any error(s) in your solutions, and/or and additional questions

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Simply copy your posted document into a text editor and insert revisions,

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A pendulum requires 2 seconds to complete a cycle, which consists of a

How long does it take to get from one extreme point to the other, how long from

A complete cycle consists of motion

• from extreme point to equilibrium

• then to opposite extreme point

• then back to equilibrium and

• finally to the original extreme point. 

The cycle can thus be broken into four parts. 

The time required for a cycle does not vary.

Making the reasonable assumption that the time required to move from

equilibrium to extreme point is the same as the time required to move from

extreme point back to equilibrium, the four parts of the cycle will all take the

same time.  If our assumption is correct, we have therefore broken the

cycle into four equal quarter-cycles.

• The time required for each quarter-cycle is 1/4 of the time required for

  a complete cycle.

In this case it takes 2 seconds to complete a cycle, so the 1/4-cycle time

is 1/4 * 2 sec = 1/2 sec or .5 sec.

• To get from one extreme point to the other requires two quarter-cycles

  (extreme to equil then equil to extreme), which would take 2 * .5 second = 1

  second.

• To get from extreme to opposite extreme and back to equilibrium takes

  another 1/4 cycle, a total of three quarter-cycles, which required 3 * .5

  second = 1.5 second.

Related notes:

You can't base an analysis of pendulum motion on the assumption of uniform

acceleration:

• As has been observed with the 'pearl pendulum' and in the introductory

  pendulum experiment, as the amplitude of a freely swinging pendulum

  decreases, the period of its motion does not change significantly.

• So the pendulum swings through decreasing distances in equal time

  intervals.

• Its average velocity during the 1/4 cycle between extreme point, where

  it is at rest, and equilibrium therefore decreases, while the time interval

  remains constant.

• Its velocity therefore changes by less and less, but during equal time

  intervals.

• Its average acceleration therefore decreases with time, contradicting an

  assumption of uniform acceleration.

For university physics students (and other interested students whose

background includes calculus):

If x(t) is the position function, then its derivative dx/dt, also denoted x

'' (t) or just x '', is the instantaneous-velocity function.

• We will see later that for a pendulum, the position, velocity and acceleration are all

  trigonometric functions of clock time. None of theses functions is linear (so, for example,

  velocity isn't a linear function of clock time, meaning that acceleration isn't

  constant).

Here's a series of key questions that outline how we know that position,

velocity and acceleration of an ideal pendulum are all trigonometric functions

of clock time. We don't 'officially' encounter them until late in the

semester, but they apply to the present situation and they can be answered using

knowledge of first-semester calculus. You don't need to answer them now but if

you want to try you're welcome:

• Can you name a function x(t) satisfies the equation x '' ( t) = - x(t)? Hint: At

  least one of the basic functions you studied in precalculus satisfies this

  equation.

• Can you now name a function x(t) satisfies the equation x '' ( t) = - 4 x(t)?

• How about the equation m * x '' ( t) = - m g / L * x(t)?

• If you have answered the questions and have a function x(t), then find

  its derivative to get the velocity function v(t) = x ' (t); and find the

  second derivative to get the acceleration function a(t) = x '' (t).