[Physics waves] Spring and Block on inclined plane

AI Thread Summary
A block weighing 9.81N oscillates on a 60-degree inclined plane attached to a spring with a constant of k=100N/m and an equilibrium length of 0.6 m. The initial kinetic energy of the block is 2J, which equals the potential energy of the spring at a specific point in its motion. The discussion highlights the importance of considering gravitational potential energy in the total energy calculations of the block-spring system. Participants debate how to account for gravitational effects on the amplitude of oscillation and the correct method to calculate the maximum length of the spring during oscillation. Ultimately, it is concluded that gravitational potential energy is integrated into the spring's potential energy calculation, leading to a successful resolution of the problem.
masterchiefo
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Homework Statement


A block weighing 9.81N oscillates on a plane inclined of 60 degrees attached to a spring whose constant
point is k=100N/m and the length is to equilibrium l0 = 0.6 m. While the block goes to the
top losing speed , a timer is started ( to t0 = 0) when the block passes through the
point where the potential energy of the spring assembly system equals the kinetic energy of the block (U0 =K0) and
measuring the K(t0)= K0 = 2J. Note that the inclined plane is also the potential energy
Total block -spring system is given by
U = 1/2* kx2 .
The block kinetic energy is
K =1/2* m v2 .
The total energy of the system is preserved and satisfies the relationship :
E = U + K =1/2 *kA2.
a) What are the initial position xo and vo initial speed of the block ?
b) What is the maximum length of the spring when it oscillates ?

Homework Equations

The Attempt at a Solution


a)
since K0 = U0
and K0 = 2J then U0 also equals 2J.

Then I can use those 2 equations:

W = 9.81N so m = 1kg

U = 1/2* k*x2 .
2J = 1/2* 100*x2 .
x= 0.2m
K =1/2* m v2
2J = 1/2 * 1kg * v2
v = 2m/s

b)
we know that l0 = 0.6
W = sqrt(k/m)
W = sqrt(100/1) = 10rad/s

E = U + K =1/2 *kA2.
E = 2J + 2J =1/2 *kA2.
2J + 2J =1/2 *100*A2.
A = 0.282843

I am pretty much stuck here and not sure how to proceed to find the maximum length of the spring when it oscillates.
 
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Why are you not including the gravitational potential energy of the block?
 
andrevdh said:
Why are you not including the gravitational potential energy of the block?
It is added to find A.
U is the potential energy
 
Anyone can help me with this?
 
Since the block is oscillating on an incline one would think that the gravitational potential energy would change as is goes up and down the incline. I thought U is the potential energy stored in the spring?
 
andrevdh said:
Since the block is oscillating on an incline one would think that the gravitational potential energy would change as is goes up and down the incline. I thought U is the potential energy stored in the spring?
Thats the part I don't understand.
How can I calculate that ?
 
I should think that you need to choose a zero level and then calculate the vertical height above or below this point with the given angle.
 
If you choose the zero level for the gravitational potential energy, UG, at the to point the the total energy of the system would be ... joule at every other point in the oscillation.
 
andrevdh said:
If you choose the zero level for the gravitational potential energy, UG, at the to point the the total energy of the system would be ... joule at every other point in the oscillation.
1/2* kx2 .+1/2* m v2 .=1/2 *kA2.

I know A,k,m
and I don't know x and v, how am I supposed to do this to get max length of spring?
 
  • #10
lo seems to be the length of the spring at equilibrium. The max length would then be when it is at the bottom of the oscillation. So it would just be lo + A. That is the block oscillates with amplitude A about the equilibrium position. Here I am assuming the block hangs from the spring rather than sitting on top of it. Problem is I don't think you calculated A correctly. It should really help you if you make a drawing.
 
  • #11
andrevdh said:
lo seems to be the length of the spring at equilibrium. The max length would then be when it is at the bottom of the oscillation. So it would just be lo + A. That is the block oscillates with amplitude A about the equilibrium position. Here I am assuming the block hangs from the spring rather than sitting on top of it. Problem is I don't think you calculated A correctly. It should really help you if you make a drawing.
How else would I calculate A?
and yes that is how it is, block hangs from the spring.
 
  • #12
I still need help :(
 
  • #13
I still need help :(
 
  • #14
Can anyone help me ?
 
  • #15
You can calculate the phase constant associated with the SHM if the zero gravitational potential is chosen at the point where the energy stored in the spring is equal to the kinetic energy of the block. Which would then enable you calculate its initial position.
 
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  • #16
You could also determine the phase at which the velocity is zero and use it to determine the displacement at that stage, which would then be the amplitude.
 
  • #17
andrevdh said:
You could also determine the phase at which the velocity is zero and use it to determine the displacement at that stage, which would then be the amplitude.
v0 = −Aω sin(ωt0 + ϕ)
0 = −Aω sin(ω*0 + ϕ)
0 = −Aω sin( ϕ)
ω = sqrt( 100/ 1) = 10
0 = −A*10 sin( ϕ)
I don't know A and ϕ, how could I possibly find 2 unknown variable ?
 
  • #18
You have assumed incorrectly that the velocity is zero at t = 0 or to. I am losing the line of reasoning since this is stretching over such a long period of time. What do you mean by this?
masterchiefo said:
...Note that the inclined plane is also the potential energy
should it be "Note that the bottom of the inclined plane is also the zero potential energy level"?
 
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  • #19
andrevdh said:
You have assumed incorrectly that the velocity is zero at t = 0 or to. I am losing the line of reasoning since this is stretching over such a long period of time. What do you mean by this?

should it be "Note that the bottom of the inclined plane is also the zero potential energy level"?
Should of said this: " The total potential energy block -spring system is given by U = 1/2* k*x2 . "

But then how do I know which "t" to use ?
What I am wondering is why the A I calculated in the beginning is wrong ?
 
  • #20
Well, I was thinking along the line that in the energy conservation game it has a certain amount of kinetic energy, 2J, at this point and it will travel upwards until it is converted into potential energy, but not only that stored in the spring, also that stored in the block-earth configuration or gravitational potential energy. As to the time I thought if you could somehow determine just the phase at which the velocity is zero, then you could use the same phase to determine the position of the oscillator, which would be its amplitude. The problem seems to boil down to how do gravity influence an shm oscillator. It does not change its frequency, but how do it factor into its amplitude?
 
  • #21
andrevdh said:
Well, I was thinking along the line that in the energy conservation game it has a certain amount of kinetic energy, 2J, at this point and it will travel upwards until it is converted into potential energy, but not only that stored in the spring, also that stored in the block-earth configuration or gravitational potential energy. As to the time I thought if you could somehow determine just the phase at which the velocity is zero, then you could use the same phase to determine the position of the oscillator, which would be its amplitude. The problem seems to boil down to how do gravity influence an shm oscillator. It does not change its frequency, but how do it factor into its amplitude?
The A I calculated was correct, I got this problem done. thanks.
 
  • #22
Its a pleasure. It seems that the gravitational potential energy is somehow incorporated into the potential energy calculation of the spring, the ½kx2 term, if you think logically about it. Also one can calculate the time at which the speed is zero by first determining the phase constant from the given information. I got the impression that that is the way they wanted you to solve the problem, but you found an easy way to solve it.
 
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