Oscillating Spring in a Frictionless pulley system

[jschim]

Homework Statement

Oscillating Spring Problem With a Pulley. Need Answer by 3/3/20 please help.

Relevant Equations

w=sqrt(k/m)
f=ma
Us=1/2kx^2

1. Draw Free body Diagram for each weight.

2. Solve for Tension in Rope.

3. Find Spring Constant.

4. Find omega (w, or angular velocity)

 

[kuruman]

Is there a question you wish to ask? We don't do your homework for you. Tell us what you think and show your attempt at a solution. Start with the free body diagrams. This might take longer than you have allowed.

 

[haruspex]

There is not enough information. E.g. consider spring constant =0; it does not contradict any given info.
Need to know the equilibrium length of the spring as well as the relaxed length. Or maybe 0.2m is supposed to be the equilibrium extension of the spring?

 

[kuruman]

There is not enough information. E.g. consider spring constant =0; it does not contradict any given info.
Need to know the equilibrium length of the spring as well as the relaxed length. Or maybe 0.2m is supposed to be the equilibrium extension of the spring?

I was about to edit my post and add that we need the full statement of the problem. The 0.2 m, a.k.a. the unstrained length, must be the no load length.

 

[jschim]

the spring at equilibrium is 0.2 meters

 

[haruspex]

I was about to edit my post and add that we need the full statement of the problem. The 0.2 m, a.k.a. the unstrained length, must be the no load length.

the spring at equilibrium is 0.2 meters

I was careful to write "equilibrium extension", i.e the extension at equilibrium. Knowing only the equilibrium length, but not the relaxed length, doesn't help.

 

[jschim]

I was careful to write "equilibrium extension", i.e the extension at equilibrium. Knowing only the equilibrium length, but not the relaxed length, doesn't help.

sorry, I am not very good at physics, the equilibrium extension is 0.2 meters

 

[haruspex]

sorry, I am not very good at physics, the equilibrium extension is 0.2 meters

Ok, so post an attempt, per forum rules.

 

[jschim]

Ok, so post an attempt, per forum rules.

here is the free body diagrams, I am working on the rest

 

[haruspex]

View attachment 258016

here is the free body diagrams, I am working on the rest

You should only consider forces acting directly on a body. Mass m1 doesn’t 'know' anything about m2g, just the tension. Similarly m2.

 

[jschim]

I tried my best, but I am not sure how to work around finding tension with the spring involved

 

[jschim]

here are the last two parts. Thank you for helping me!

 

[jschim]

You should only consider forces acting directly on a body. Mass m1 doesn’t 'know' anything about m2g, just the tension. Similarly m2.

I tried part two again based on you advice and I think I did it correctly.

Fnet=ma

T-mg=mg

T=2mg

T=2(.4)(9.8)

T=7.84 N

 

[jschim]

View attachment 258018View attachment 258019
here are the last two parts. Thank you for helping me!

I just realized I messed up my units on #3 by a power of 10. It should be .588 rather than .0588 and 470.4 rather than 47.4

 

[haruspex]

T-mg=mg

I assume this is for m2.
How many forces act on m2 and what are they?

 

[jschim]

I assume this is for m2.
How many forces act on m2 and what are they?

weight acts downwards, and tension acts upwards

 

[haruspex]

weight acts downwards, and tension acts upwards

Right, so what equation does that give for balance?

 

[jschim]

Right, so what equation does that give for balance?

netF=tension-weight

right??

 

[kuruman]

netF=tension-weight

right??

Right, this is the net force on the hanging mass. What is it equal to?

 

[jschim]

Right, this is the net force on the hanging mass. What is it equal to?

It would be equal to the hanging mass times acceleration, in this case, gravity.

f=ma

 

[kuruman]

Correct. Can you write another equation expressing Newton's 2nd law using the free body diagram of the mass attached to the spring? Do that using symbols, not numbers. Also to distinguish the masses from each other use m1 for the mass on the table and m2 for the hanging mass. Once you have written the second equation, put them on two separate lines, one below the other.

 

[haruspex]

It would be equal to the hanging mass times acceleration

Yes.

in this case, gravity.

It is not accelerating. It is stationary until the spring is released.

 

[haruspex]

Correct

No, what @jschim wrote in post #20 goes wrong at the end. I suspect this is how he or she got the erroneous T-mg=mg in post #13.

 

[kuruman]

No, what @jschim wrote in post #20 goes wrong at the end. I suspect this is how he or she got the erroneous T-mg=mg in post #13.

Yes, of course. I guess I stopped reading after "mass times acceleration".