Understanding SHM: Solving Difficult Homework Problems

[PhysicStud01]

Homework Statement

Homework Equations

the equation is given in the question

The Attempt at a Solution

i tried replacing the 2/3 pi rad as the cos argument and i obtained 2. but the mark scheme says that the position is 3cm above AB.

in the next part, we have a phase diff of 4/3 pi and i again obtained 2 but the MS still says 3cm above AB. can someone help. am i doing a mistake here.

 

[BvU]

Hello PS,

Could you type out the essential, concise, complete problem statement and select which of the equations you want to use in solvig it / And then post your work, please ?

My magnifying glass doesn't allow me to read the small print, and besides, it would take away too much time away from helping others.

 

[PhysicStud01]

Hello PS,

Could you type out the essential, concise, complete problem statement and select which of the equations you want to use in solvig it / And then post your work, please ?

My magnifying glass doesn't allow me to read the small print, and besides, it would take away too much time away from helping others.

could yiou try dragging the picture into another tab and enlarge it.
i uploaded it with an appropriate size but it's not appearing correctly here.

it's question 3 part b at
http://www.docstoc.com/docs/173453412/9702_w10_qp_43

 

[BvU]

And what is the problem statement when you reproduce it ? (see the guidelines: it helps you too -- to focus and order your thinking)

 

[PhysicStud01]

And what is the problem statement when you reproduce it ? (see the guidelines: it helps you too -- to focus and order your thinking)

i could work out the in-between questions.
for the part I'm having problems, i replaced the angle given in the equation given in the question and i obtained 2. the answer says it's 3cb above AB.

i checked again and there don't seem to be a problem with my solution, in my opinion

 

[BvU]

See the guidelines. O&O.

 

[PhysicStud01]

i did it and still can't understand the problem

 

[andrevdh]

Make a rough drawing of the displacement - time graph of X and post it please.

 

[PhysicStud01]

Make a rough drawing of the displacement - time graph of X and post it please.

i already understand how the graph is. its a -cos with height 4

 

[andrevdh]

How would the graph look with respect to the previous one if it was leading with say π radians?
That is the phase difference between the two is π radians, with the 2nd one leading?
Maybe draw in pencil and post the photo.
This might be clearer to you:

 

[PhysicStud01]

How would the graph look with respect to the previous one if it was leading with say π radians?
That is the phase difference between the two is π radians, with the 2nd one leading?
Maybe draw in pencil and post the photo.

in red.

also, i don't know how to do it by calculation. i tried but did not get the asnwer as i said previously. try to help with this too

 

[andrevdh]

Look at the photo in my previous post.
Do that make any sense to you?

 

[PhysicStud01]

Look at the photo in my previous post.
Do that make any sense to you?

could you relate it to the type of graph i drew please

 

[andrevdh]

The radius line is rotating at a constant speed anti-clockwise.
It projection on the y-axis is the displacement of the piston.
At time zero X is at the bottom.

 

[PhysicStud01]

The radius line is rotating at a constant speed anti-clockwise.
It projection on the y-axis is the displacement of the piston.
At time zero X is at the bottom.

yeah, but how does that make the position of Y be 3cm above AB.
and the same is obtained if the angle is 4pi / 3

could do it it by calculation please. this part actually scores only 1 mark, so i think it should be very simple but i can't get it.

 

[andrevdh]

The phase difference between the two is Φ = (2π)/3 rad.
That is the meaning of the other two lines.

 

[PhysicStud01]

The phase difference between the two is Φ = (2π)/3 rad.
That is the meaning of the other two lines.

but how do i obtained the distance traveled / position of the piston?
AB and CD are separated by 8cm. the amplitude of the oscillation is 4.0cm.

when phase difference is pi, when one is at AB, the other is at CD, right?

 

[andrevdh]

That would be if the phase difference was π, which it is not.

 

[PhysicStud01]

That would be if the phase difference was π, which it is not.

yeah, but its 120, it's greater than 90, so, it should be greater than 4cm and it's less than 180, so, the distance should be less than 8cmcould you tell me how to do this by calculation please. i really need to understand this quick

 

[andrevdh]

Yes, so the radius line is 120 degrees before that of X, that is Y is at the two o'clock position in my drawing.

 

[PhysicStud01]

Yes, so the radius line is 120 degrees before that of X, that is Y is at the two o'clock position in my drawing.

but how do i identify the position form this.
i used the formula given in the question, but it's not giving the correct answer

 

[andrevdh]

The general formula looks like this to compensate for a phase difference, Φ,

y = y_m cos(ωt+Φ)

or the sine function is used. In this case the phase difference is 2π/3

 

[PhysicStud01]

The general formula looks like this to compensate for a phase difference, Φ,

y = y_m cos(ωt+Φ)

or the sine function is used. In this case the phase difference is 2π/3

i used it but obtianed 2 instead of 3

i put t = 0

 

[andrevdh]

Yes, that is what I get too.

 

[PhysicStud01]

Yes, that is what I get too.

has anyone tried it too?

plus, does it not look wild for a phase difference of 120 to be only 2 or 3 when a phase difference of 180 correponds to a distance of 8.0cm

 

[PhysicStud01]

additoinally, I've been thinking about this. the displacement obtained from the equation, it's from the equilibrium position, right? not from AB. then, the 2.0cm obtained is actually from equilibrium position, and 4+2 = 6cm from AB?