How Does Adding Weight Affect the Oscillation of a Board on Rollers?

[Maneuver1]

Homework Statement

The following problem is taken from the MCAT Prep book ExamKrackers 1001 Physics Problems

A board with mass m and center of mass C sits on fixed, rotating rollers separated by a distance L. The coefficient of friction between the board and the rollers is u. The board oscillates back and forth in simple harmonic motion at a frequency f with the center of gravity moving a distance d from the center of the rollers. The period of motion T is given by the equation above the figure.

If a weight were placed on top of the board, which of the following would occur?

A. Both f and d would decrease
B. Both f and d would increase
C. Both f and d would remain the same
D. d would decrease and f would remain the same.

Homework Equations

The given equation above.

The Attempt at a Solution

To be honest I have no idea how to approach this question. From the equation above it would seem that period and therefore frequency does not depend on mass. Therefore this leaves the possible answer choices to options C and D. Assuming that d is like the amplitude I would think that increasing the normal force on the rollers by adding a mass would increase the restoring Force which would increase the acceleration and therefore the amplitude. Any insight would be appreciated.

 

[kuruman]

On the basis of what you know about harmonic motion, if the mass increases, would the frequency increase, decrease, or stay the same? Think, for example, of a spring-mass system and assume that the board is attached to a spring or any device that produces a spring-like restoring force.

 

[cartonn30gel]

I think one should use the given equation in the problem since this may not necessarily be the same as a mass-spring system. Note that mass doesn't appear in that equation whereas it does in the equation governing regular mass-spring system. I suggest sticking to
$$T=2\pi \sqrt{\frac{L}{2\mu g}}$$
and considering how those variables change as you add that extra mass.

 

[kuruman]

I think one should use the given equation in the problem since this may not necessarily be the same as a mass-spring system. Note that mass doesn't appear in that equation whereas it does in the equation governing regular mass-spring system. I suggest sticking to
$$T=2\pi \sqrt{\frac{L}{2\mu g}}$$
and considering how those variables change as you add that extra mass.

Indeed the problem may not necessarily be a spring-mass system. However, the fact remains that for a mass undergoing one-dimensional simple harmonic motion, the restoring force on the mass is proportional to the displacement. Don't call the proportionality factor k, call it β. Then

F = -βx = ma, from which you get a = -(β/m)x = -ω²x

This makes clear what happens to the frequency when you add mass to the system.

What happens to the amplitude d is a whole 'nother issue. If the mass is added when the oscillator is at maximum displacement, you get a different answer from if the mass is added at some other point of the motion.

 

[aim1732]

Hmmm. There is very common linear simple harmonic system whose frequency does not depend on mass - the simple pendulum.

 

[kuruman]

Hmmm. There is very common linear simple harmonic system whose frequency does not depend on mass - the simple pendulum.

I agree, but this case is simple harmonic motion in a straight line.

 

[aim1732]

I don't see your point, sorry. The only effect the mass has is a frictional force that, as long as ma>μmg, this friction will be ma=+mω²x and if this is put in the equation that provides the time period:
-μ(M+m)g * 4x/2l + mω²x = -Mω²x
The mass as you see will cancel.

 

[kuruman]

I don't see your point, sorry. The only effect the mass has is a frictional force that, as long as ma>μmg, this friction will be ma=+mω²x and if this is put in the equation that provides the time period:
-μ(M+m)g * 4x/2l + mω²x = -Mω²x
The mass as you see will cancel.

Sorry, I don't see how you get this equation. Maybe I am misunderstanding the problem. How would you write Newton's Second Law in this case, with and without the extra mass?

 

[aim1732]

Sorry but I was thinking it would be unwise to solve the problem on someone else's thread.
There are two steps - one to write out Newton's Second law in th vertical direction and the equivalent law of rotation - taking torques about centre of mass that is - to obtain normal forces the rollers exert on the board. Then assuming this friction has it's maximum value multiplication with μ yields friction component. The net force in the horizontal direction is then comes out to be proportional to x.
One more of my assumptions is that the mass is placed directly above the centre of mass so it exerts no torque about it.

 

[kuruman]

Sorry but I was thinking it would be unwise to solve the problem on someone else's thread.

I agree. This is my last posting on this thread.

There are two steps - one to write out Newton's Second law in th vertical direction and the equivalent law of rotation - taking torques about centre of mass that is - to obtain normal forces the rollers exert on the board. Then assuming this friction has it's maximum value multiplication with μ yields friction component.

That's my problem with this problem. What if the friction required has not reached its maximum value? There is no language in the problem saying that it has. I believe this is an unwarranted assumption, but I understand where you are coming from. It is the only way to introduce μ in the picture.

To Maneuver1 (original poster): If you are confused by all this, my apologies. MCAT problems are sometimes ambiguous, but I wish you success when you take the test.

 

[aim1732]

Being a student makes us a problem solver and it won't be any use banging our heads over an obvious assumption (which is not an approximation so the solution is technically wrong). But honestly I think I should have mentioned before.

 

[Maneuver1]

Thanks for the help. This problem is poorly worded. I appreciate the insight