cq1_10_1 solution and discussion

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 or comments you might have.

Simply copy your posted document into a text editor and insert revisions, questions, and/or self-critiques, marking your insertions with ####. Submit using the Submit Work Form.

A pendulum requires 2 seconds to complete a cycle, which consists of a complete back-and-forth oscillation (extreme
point to equilibrium to opposite extreme point back to equilibrium and finally to the original extreme point). As long as the
amplitude of the motion (the amplitude is the distance from the equilibrium position to the extreme point) is small
compared to the length of the pendulum, the time required for a cycle is independent of the amplitude.

How long does it take to get from one extreme point to the other, how long from an extreme point to equilibrium, and
how long to go from extreme point to equilibrium to opposite extreme point and back to equilibrium?

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).

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 or comments you might have.

Simply copy your posted document into a text editor and insert revisions, questions, and/or self-critiques, marking your insertions with ####. Submit using the Submit Work Form.

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.

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).

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).