Equilibrium of Two Masses Connected by a Spring: Find Velocity and Distance

  • Thread starter DDS
  • Start date
  • Tags
    Stuck
In summary: I'll wait 1/4 second to be sure...)In summary, a mass m1 = 9.23 kg is connected to a light spring of constant k = 106 N/m and a second mass m2 = 6.42 kg is pushed against m1, compressing the spring by 0.169 m. When the system is released, both masses move to the right on a frictionless surface. m1 reaches equilibrium and m2 loses contact, moving to the right with speed v = 0.439 m/s. To find the distance between the masses when the spring is fully stretched for the first time, the period and time of m1's oscillation must be determined. The period is 1.85
  • #1
DDS
171
0
A mass, m1 = 9.23 kg, is in equilibrium while connected to a light spring of constant k = 106 N/m that is fastened to a wall.A second mass, m2 = 6.42 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.169 mThe system is then released, and both masses start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 and moves to the right with speed v. A) Determine the value of v.B)How far apart are the masses when the spring is fully stretched for the first time.

I have solved part A by using :
1/2KA^2=1/2mv^2
v=0.439 m/s

for part b i know i have to find a new amplitude for block m1 and i did so.

1/2KA^2=1/2mv^2
A=0.1295 m

Here is where i am stuck, the amplitude i found tells me the furthest distance that block m1 travels. However everytime i try to solve for time and then distance i get the wrong answer. Can someone help me find time and distance thus solving this problem??
 
Physics news on Phys.org
  • #2
DDS said:
Here is where i am stuck, the amplitude i found tells me the furthest distance that block m1 travels. However everytime i try to solve for time and then distance i get the wrong answer. Can someone help me find time and distance thus solving this problem??
Hint: Once m2 leaves contact, m1 will execute Simple Harmonic Motion. What's the period of its motion?
 
Last edited:
  • #3
I know that m1 will undergo SHM, i have already gone about this route and got stuck so i back tracked and tried the apporach i posted. But what i did this way was:

w= sqrt k/m
w=3.3888

thus
w=pi/T
T=0.926
 
  • #4
DDS said:
I know that m1 will undergo SHM, i have already gone about this route and got stuck so i back tracked and tried the apporach i posted.
What you posted above (in your first post) is a necessary part of the solution. But it's not complete, since it makes no mention of time.

But what i did this way was:

w= sqrt k/m
w=3.3888

thus
w=pi/T
T=0.926
[tex]T = 2 \pi \sqrt{\frac{m}{k}}[/tex]

Now use your knowledge of the period to find the time it takes for m1 to go from the equilibrium point to its maximum extension. (What fraction of the period is this?) Once you know the time you can figure out how far m2 traveled in that time.
 
  • #5
I know this is one quarter of the period but i am not exactly 100% sure how to manipulate my period inorder to find time. This is where i got stuck in the original post.
 
  • #6
Recheck your equation w = sqrt k/m you are missing a 2 in there. So when you find the period of the mass(a) and spring system you can determine the time for the mass (a) to reach its max amplitude and then determine how far mass (b) has traveled relative to mass (a). If you haven't already then draw it out, pictures can be very helpful.
 
  • #7
I forgot to post this above but my period is 1.85
 
  • #8
Period is time/cycle. The units are often expressed as time only
 
  • #9
DDS said:
I know this is one quarter of the period but i am not exactly 100% sure how to manipulate my period inorder to find time. This is where i got stuck in the original post.
Right. The period is the time for one complete cycle of motion. (If you used standard units, the period will be given in seconds.)
 
  • #10
Silly question but in order to find my cycles would i find my frequency then take that answer and multiply it by speed thus giving me:

F=1/T
F=0.54*0.439
cycles per second =0.237
 
  • #11
This is what i am trying to do. I have my Period which is 1.85 , now i know that period is time/cycles per second so i am trying to find my time in which it takes m1 to travle its max distance.

I figured that F=1/T=0.5405 thus 1.85*0.5405= Time it takes m1 to travel its max distance. Am i right?
 
  • #12
DDS said:
Silly question but in order to find my cycles would i find my frequency then take that answer and multiply it by speed thus giving me:

F=1/T
F=0.54*0.439
cycles per second =0.237
Not sure what you are doing here. Looks like you found the frequency and multiplied this by the speed, for some reason. Check your units! (1/s)(m/s) = (m/s^2).

For this problem, all you need is the period. If you wish to find the frequency (number of cycles per second) at which m1 will oscillate, then use f = 1/T. (Don't know what possessed you to "multiply by speed".)
 
  • #13
no no that's wrong...hmmm see i am not sure what to due with my period to find time
 
  • #14
DDS said:
This is what i am trying to do. I have my Period which is 1.85 , now i know that period is time/cycles per second so i am trying to find my time in which it takes m1 to travle its max distance.

I figured that F=1/T=0.5405 thus 1.85*0.5405= Time it takes m1 to travel its max distance. Am i right?
Nope! Does it even make sense? This time you are multiplying frequency x period. What answer do you think you'll get? [itex]f \times T = (1/T) \times T = 1[/itex].

This is much easier than you think. You've stated the answer yourself in and earlier post
I know this is one quarter of the period...
 
  • #15
pete worthington said:
So when you find the period of the mass(a) and spring system you can determine the time for the mass (a) to reach its max amplitude and then determine how far mass (b) has traveled relative to mass (a).

I found my period but i am stuck on how to find time using my period
 
  • #16
Let's rephrase the question:

How much time does it take for m1 to travel 1/4 of an oscillation?
 
  • #17
Would it be 0.4625 seconds?
 
  • #18
DDS said:
Would it be 0.4625 seconds?
Of course. (Now please smack yourself in the head for me. Just kidding.)

Now... how far does each mass get in that time?
 
  • #19
LOL i think i really do deserve a smack, i was confusing myself with the units of period and that threw me off completely. I figured out my answer. Thank you to everyone for your help.

However i do have one quick question, i had a little bit of trouble determining that this is 1/4 of a cycle. Can some clarify my doubts about how to determine this value in the future. I wouldn't care if you use my problem as a example.
 
  • #20
DDS said:
However i do have one quick question, i had a little bit of trouble determining that this is 1/4 of a cycle. Can some clarify my doubts about how to determine this value in the future. I wouldn't care if you use my problem as a example.
I suggest that you plot the position of m1 as a function of time, starting from the equilibrium point. (You don't need to be super accurate; just get the basic idea.) It should look like a sine function. That may help you.
 
  • #21
That still leaves things kind of blury i know that in SHM its either a sin or cos function that gets manipulated my pi/2 to get the other function but in this case when i found that its a 1/4 cycle i found it by seting up my inital conditions and then finding the phase constant. Is there a simpler way??
 
  • #22
If you're a kinesthetic learner I would get a low "k" spring (any hardware store), hang it from the ceiling and attach mass to the bottom of it. Set it into motion and identify the amplitude, the equillibrium position, the period etc. You can also use a rubber band. This usually drives home the relationship between Period, amplitude and time for 1/4 cycle. Cosine is when you release mass from rest at the max amplitude. Sine is when the mass is at rest at equillibrium and suddenly accelerated .with an impulse.
 

FAQ: Equilibrium of Two Masses Connected by a Spring: Find Velocity and Distance

What is the definition of equilibrium in the context of two masses connected by a spring?

Equilibrium in this context refers to a state where the forces acting on the two masses are equal and opposite, resulting in a stable and balanced system.

How do you calculate the velocity of the two masses in this system?

The velocity of the two masses can be calculated using the equation v = sqrt(k/m), where k is the spring constant and m is the mass of the system.

What is the equation for finding the distance between the two masses in equilibrium?

The equation for finding the distance between the two masses in equilibrium is x = F/k, where F is the force applied to the spring and k is the spring constant.

How does the spring constant affect the equilibrium of the system?

The spring constant determines the stiffness of the spring and affects the force required to stretch or compress the spring. A higher spring constant will result in a smaller distance between the two masses in equilibrium.

Can equilibrium be achieved if the two masses have unequal masses?

Yes, equilibrium can still be achieved if the two masses have unequal masses as long as the spring constant is adjusted accordingly to balance out the forces acting on the system.

Back
Top