QA 35

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course PHY 121

8/2 5

035. Velocity and Energy in SHM

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Question: `q001. Note that this assignment contains 5 questions.

At its maximum velocity, a simple harmonic oscillator matches the speed of the point

moving around its reference circle. What is the maximum velocity of a pendulum 20 cm

long whose amplitude of oscillation is 2 cm? Note that the radius of the reference

circle is equal to the amplitude of oscillation.

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Your solution:

period of a pendulum: .2 `sqrt L

T = .2 `sqrt 20 cm

T = .2 (4.4721)

T = .89 s

`dt = 2 `pi / `omega

`omega = 2 `pi/`dt

`omega = 2 `pi / .89 s

`omega = 7.06 radians/s

confidence rating #$&*:

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Given Solution:

We need to find the velocity of the point on the reference circle that models this

motion. The reference circle will have a radius that matches the amplitude of

oscillation, in this case 20 cm. The period of the oscillation is T = 2 `pi `sqrt( L /

g ) = 2 `pi `sqrt( 20 cm / (980 cm/s^2) ) = .9 sec, approx..

Thus the point completes a revolution around the reference circle once every .9 sec.

The circumference of the reference circle is 2 `pi r = 2 `pi * 2 cm = 12.6 cm, approx.,

so the point moves at an average speed of 12.6 cm / .9 sec = 14 cm/s.

Thus the maximum velocity of the pendulum must be 14 cm/s.

STUDENT COMMENT:

i got the right answer, but went about it a different way than you

INSTRUCTOR RESPONSE:

You used the formula vMax = omega*A, which is fine. However that formula is a direct

result of the reference-circle model and easiest to understand in terms of that model.

omega * A is the speed of the point on the reference circle, and the maximum speed of

the oscillating object occurs when the direction of motion of the reference-circle point

matches that of the object.

A little more detail in the description:

At the extreme points of the pendulum's motion, the reference-circle point is moving at

a right angle to the pendulum's path. The pendulum's position 'shadows' that of the

reference circle so at that instant the pendulum is stationary.

When the pendulum passes through equilibrium the reference circle point is moving in

exactly the same direction as the pendulum, so the pendulum's speed is equal to that of

the reference circle point.

At every other point the pendulum is moving, but not as fast as the reference circle

point.

So the maximum pendulum speed is equal to that of the reference circle point, and occurs

as the pendulum passes through the equilibrium position.

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Self-critique (if necessary):

In the first part, I believe that I took a short cut that I found in my notes. The

second part was a matter of neglecting to multiply the radius.

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Self-critique rating: 2

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Question: `q002. If the 10 kg mass suspended from the 3000 N/m spring undergoes SHM

with amplitude 3 cm, what is its maximum velocity?

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Your solution:

`omega = `sqrt k/m

`omega = `sqrt 3000 N/m / 10 kg

`omega = `sqrt 300

`omega - 17.32 radians /s

Just for the record, I'm sure this is wrong. I did not use the 3 cm amplitude/radius

information. I also figured out that a 10 kg mass will have a force of 98 Newtons,

which I did not use, but might have needed.

confidence rating #$&*:

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Given Solution:

We previously found the angular frequency and then the period of this system, obtaining

period of oscillation T = .36 second. The reference circle will have radius 3 cm, so

its circumference is 2 `pi * 3 cm = 19 cm, approx..

Traveling 19 cm in .36 sec the speed of the point on the reference circle is

approximately 19 cm / (.36 sec) = 55 cm/s. This is the maximum velocity of the

oscillator.

STUDENT COMMENT:

i figured the period right but didn't document it, i didn't think to use the

circumferenc to get the distance, i used the

equations from the previous question, what you have written for the answer seems to be v

avg

INSTRUCTOR RESPONSE:

That result is vAve for the speed of the reference-circle point; it was perceptive of

you to notice that.

In the reference-circle model the reference points moves with constant speed, and

therefore with constant angular velocity.

The maximum speed of the oscillating object is the speed of the reference-circle point,

and occurs when the object is moving in the same direction as the reference-circle

point.

STUDENT COMMENT: i didn't think to use the circumference to get the distance, i used

the equations from the previous question, what you have written for the answer seems to

be v avg

INSTRUCTOR RESPONSE:

That is the average speed of the point moving around the reference circle. In general

this speed is omega * r.

However in some positions the reference-circle point is moving perpendicular to the axis

along which the actual object is moving, and at these points the speed of the object is

zero.

At only two points is the velocity of the reference-circle point parallel to the motion

of the object, so only at these two points is the speed of the object equal to that of

the reference-circle point.

At all other points the velocity of the reference-circle point is neither parallel nor

perpendicular to that of the object, so the speed of the object is neither zero nor

equal to that of the reference-circle point. At these points the object's speed is in

between zero and the speed of the reference point, and the object's velocity is omega *

A * cos(omega * t). Note that the cosine function has a magnitude that cannot exceed 1,

so that the speed of the object cannot exceed omega * A.

STUDENT QUESTION

I’m still getting my answers a different way. Is my solution incorrect, or is it

acceptable??

INSTRUCTOR RESPONSE

You are relying on formulas, as opposed to the unifying image behind the formulas. I

recommend the latter, just to be sure you've got the formulas right and your work makes

sense. However as long as you have the formulas right, as you generally do, your

solutions are fine.

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Self-critique (if necessary):

I didn't realize it was building off another problem. The explanation makes sense.

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Self-critique rating: 2

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Question: `q003. What is the KE of the oscillator at this speed?

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Your solution:

KE = .5 m v^2

KE = .5 (10 kg) (.55 m/s)^2

KE = 1.5125 Joules

confidence rating #$&*:

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Given Solution:

The KE is .5 m v^2 = .5 * 10 kg * (.55 m/s)^2 = 1.5 Joules, approx.. Note that this is

the maximum KE of the oscillator.

STUDENT COMMENT:

i didn't think to use the previous equation for KE from earlier in the class, i am

trying to use the new equations from the

current notes and this is throwing me off

INSTRUCTOR RESPONSE

This entire topic is difficult to sort out, and a degree of confusion at this point is

perfectly normal. You are doing well with this topic.

However it's important to understand that nothing in the current notes contradicts or

supplants the fact that the KE of mass m moving with speed v is 1/2 m v^2, and this

relationship is very important in this analysis.

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Self-critique (if necessary): OK

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Self-critique rating: 2

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Question: `q004. How much work is required to displace the mass 3 cm from its

equilibrium position?

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Your solution:

F = m * a

F = 10 kg * 9.8 m/s^2

F = 98 Newtons

`dw = F * `ds

`dw = 98 Newtons * .03 m

`dw = 2.94 Joules

confidence rating #$&*:

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Given Solution:

The mass rests at its equilibrium position so at that position there is no displacing

force, since equilibrium is the position taken in the absence of displacing forces. As

it is pulled from its equilibrium position more and more force is required, until at the

3 cm position the force is F = k x = 3000 N / m * .03 m = 90 N. (Note that F here is

not the force exerted by the spring, but the force exerted against the spring to

stretch it, so we use kx instead of -kx).

Thus the displacing force increases from 0 at equilibrium to 90 N at 3 cm from

equilibrium, and the average force exerted over the 3 cm displacement is (0 N + 90 N )

/ 2 = 45 N.

The work done by this force is `dW = F `dx = 45 N * .03 m = 1.5 Joules.

STUDENT COMMENT

My units came out to be negative. But that was only because I used F = -k * x.

INSTRUCTOR RESPONSE

The sign of the work is a common point of confusion.

When an object is displaced (at constant velocity so no KE change is involved) by means

of an applied force, against a restoring force, the applied force and the restoring

force do equal and opposite work.

F = - k x is the force exerted to restore the mass to the equilibrium position. This

conservative force does negative work, since it is directed opposite the displacement x.

By the definition of `dPE, the work done by this force is equal and opposite to the

positive change in the PE of the system.

The force required to move the mass away from equilibrium is equal and opposite to the

restoring force, is therefore in the same direction as the displacement and does

positive work on the system.

In this case the restoring force does -1.5 Joules of work, which changes the PE by - (-

1.5 Joules) = +1.5 Joules, and the force which moves the mass from equilibrium to this

position does +1.5 Joules of work.

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Self-critique (if necessary):

I obviously got mixed up on what to use for the force. Also, when I multiply 45 * .03,

I keep getting 1.35 rather than 1.5.

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Self-critique rating: 2

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Question: `q005. How does the work required to displace the mass 3 cm from its

equilibrium position compare to the maximum KE of the oscillator, which occurs at its

equilibrium position? How does this result illustrate the conservation of energy?

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Your solution:

They match in the ""Given Solutions."" I think that they show the conservation of energy

because in these cases with pendulums, springs, etc, there is a flux between potential

energy and kinetic energy. (I don't even really know what an oscillator is.) As they

move away from their equilibrium position, they are converting kinetic energy to

potential energy because they are slowing down. At the extremes of their swings, they

have no kinetic energy, for a moment, and all potential. As it moves back toward

equilibrium, the potential converts to kinetic.

confidence rating #$&*:

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Given Solution:

Both results were 1.5 Joules.

The work required to displace the oscillator to its extreme position is equal to the

maximum kinetic energy of the oscillator, which occurs at the equilibrium position. So

1.5 Joules of work must be done against the restoring force to move the oscillator from

its equilibrium position to its extreme position. When released, the oscillator

returns to its equilibrium position with that 1.5 Joules of energy in the form of

kinetic energy.

Thus the work done against the restoring force is present at the extreme position in

the form of potential energy, which is regained as the mass returns to its equilibrium

position. This kinetic energy will then be progressively lost as the oscillator moves

to its extreme position on the other side of equilibrium, at which point the system will

again have 1.5 Joules of potential energy, and the cycle will continue. At every point

between equilibrium and extreme position the total of the KE and the PE will in fact be

1.5 Joules, because whatever is lost by one form of energy is gained by the other.

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Self-critique (if necessary): OK

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Self-critique rating: 2

If you understand the assignment and were able to solve the previously given problems

from your worksheets, you should be able to complete most of the following problems

quickly and easily. If you experience difficulty with some of these problems, you will

be given notes and we will work to resolve difficulties.

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Question: `q006. A 5 kg mass is suspended from a spring whose force constant is 592 N

/ m. The system is then pulled down an additional 10 cm and released, with the result

that the mass undergoes simple harmonic motion.

What is the radius of the reference circle that models the motion?

What is the angular velocity of the reference point as it moves around this circle?

At what points of its motion will the 5 kg mass match the velocity of the reference

point?

What will be the KE of the 5 kg mass when its velocity matches that of the reference

point?

How much work does it take to pull that spring down 10 cm below its equilibrium

position?

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Your Solution:

The only thing I get out of all this is:

`omega = `sqrt k/m

`omega = `sqrt 592 N/m / 5 kg

`omega = `sqrt 118.4

`omega = 10.881 radians/second

@&

Not bad.

The object will also start out 10 cm below equilibrium, so it will oscillate between this point and 10 cm above equilibrium.

This oscillation can be modeled by a point moving around a circle of radius 10 cm, moving at omega = 10.88 rad / sec.

The object will at every instant be exactly as high (or low) as the point on the circle.

The point on the circle would be moving at 10 cm * 10.88 rad / s = 108.8 cm/second or 1.088 meters/second.

The force to pull the object 10 cm from equilibrium would be F = 10 cm * 592 N/m = .1 m * 592 N/m = 59.2 N.

The force required to pull the object down would start at 0 when the object is at equilibrium, so it would change linearly with distance from 0 N to 59.2 N, thereby averaging 29.6 N. This average force acting through 0.1 meter does work `dW = 29.6 N * 0.1 m = 2.96 Joules.

*@

confidence rating #$&*:5

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Self-critique (if necessary):

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Self-critique rating:"

&#Good responses. See my notes and let me know if you have questions. &#