# Mass-spring system on a ramp

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Thread starter 1 month ago
#1
A frictionless ramp is inclined 30 degrees to the horizontal. A massless spring is attached to a wall and it runs parallel to the line of greatest slope of the ramp down to a mass of 3000 g that rests on the ramp. The block is in equilibrium on the ramp, and the length of the spring is 120cm. The unstretched length of the spring is 80 cm. The block is moved 30 cm down the ramp and released from rest.

It says to find amplitude, period and maximum speed.

I did this
ma = kx - mgsin(theta)
ma = kx - mgsin(theta)
a = (k/m)x - gsin(theta)

so w^2 = k/m
so T = 2 * pi * sqrt(m/k)

Amplitude is 30 cm? (no calculation, just using the fact it was displaced 30 cm? Is there calculation to justify this?)

And maximum speed is wA so 0.3 * omega (in rad/s)

then substitute the values to evaluate. Is this a right method?
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1 month ago
#2
(Original post by Minibus)
A frictionless ramp is inclined 30 degrees to the horizontal. A massless spring is attached to a wall and it runs parallel to the line of greatest slope of the ramp down to a mass of 3000 g that rests on the ramp. The block is in equilibrium on the ramp, and the length of the spring is 120cm. The unstretched length of the spring is 80 cm. The block is moved 30 cm down the ramp and released from rest.

It says to find amplitude, period and maximum speed.

I did this
ma = kx - mgsin(theta)
ma = kx - mgsin(theta)
a = (k/m)x - gsin(theta)

so w^2 = k/m
so T = 2 * pi * sqrt(m/k)

Amplitude is 30 cm? (no calculation, just using the fact it was displaced 30 cm? Is there calculation to justify this?)

And maximum speed is wA so 0.3 * omega (in rad/s)

then substitute the values to evaluate. Is this a right method?
Looks sort of ok with the right diagram and justification for the steps. However, you're "guessing" the forces / ode which gives SHM and its not correct. Work through the steps methodically and upload your full working and the original question pls.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#3
(Original post by mqb2766)
Looks sort of ok with the right diagram and justification for the steps. However, you're "guessing" the forces / ode which gives SHM and its not correct. Work through the steps methodically and upload your full working pls.
I am not sure what is wrong. Here is my working:

No diagram was given with question
https://imgur.com/a/oqRCuPj
F = ma
ma = T - mgsin(theta)
T = kx
so ma = kx - mgsin(theta)
a = (k/m)x - gsin(theta)

using equilibrium point:
mgsin(theta) = k(1.2-0.8)
3 * g * sin(30) = 0.4 k
0.4 k = 1.5g
so k = 1.5g/0.4 = 15g/4

omega^2 = w^2 = k/m = 15g/(4*3) = 5g/4

so omega = sqrt(5g/4)

maximum speed = omega * amplitude = 0.3 sqrt(5g/4) m/s
period = 2pi sqrt(4/(5g)) s
amplitude = 0.3 m
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1 month ago
#4
(Original post by Minibus)
I am not sure what is wrong. Here is my working:

No diagram was given with question
https://imgur.com/a/oqRCuPj
F = ma
ma = T - mgsin(theta)
T = kx
so ma = kx - mgsin(theta)
a = (k/m)x - gsin(theta)

using equilibrium point:
mgsin(theta) = k(1.2-0.8)
3 * g * sin(30) = 0.4 k
0.4 k = 1.5g
so k = 1.5g/0.4 = 15g/4

omega^2 = w^2 = k/m = 15g/(4*3) = 5g/4

so omega = sqrt(5g/4)

maximum speed = omega * amplitude = 0.3 sqrt(5g/4) m/s
period = 2pi sqrt(4/(5g)) s
amplitude = 0.3 m
That ODE does not necessarily have a sinusoidal solution? The basic SHM equation is:
a ~ -x
So the acceleration is proportional to the negative displacement about the equilibrium point. It would help to see the original question.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#5
(Original post by mqb2766)
That ODE does not necessarily have a sinusoidal solution? The basic SHM equation is:
a ~ -x
So the acceleration is proportional to the negative displacement about the equilibrium point.
I did it wrong ? :/
Is it ma = -(T - mgsin(theta)) = mgsin(theta) - T ?
that gives a = gsin(theta) - (k/m) x

But applying N2L gives equation with opposite sign?
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1 month ago
#6
(Original post by Minibus)
I did it wrong ? :/
Is it ma = -(T - mgsin(theta)) = mgsin(theta) - T ?
that gives a = gsin(theta) - (k/m) x

But applying N2L gives equation with opposite sign?
You;ve not necessarily got the forces right. How is x defined ... and last time, pls upload the question.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#7
(Original post by mqb2766)
You;ve not necessarily got the forces right. How is x defined ... and last time, pls upload the question.
https://imgur.com/a/C6MBYeJ
[But 3000g, not 300g]

It doesn't give diagram or define x. I used x in my solution though
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1 month ago
#8
(Original post by Minibus)
https://imgur.com/a/C6MBYeJ
[But 3000g, not 300g]

It doesn't give diagram or define x. I used x in my solution though
Assuming you understand the issues with how x is defined (you have not said how you defined it) and why your ode appears to have the wrong sign for x for SHM, then for this question, wherever it comes from, it doesn't really matter as youre not marked on that part of the question. So ...

Edited ... you have used the equilibrium to get k.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#9
(Original post by mqb2766)
Assuming you understand the issues with how x is defined (you have not said how you defined it) and why your ode appears to have the wrong sign for x for SHM, then for this question, wherever it comes from, it doesn't really matter as youre not marked on that part of the question. So ...

Also note youve ben given the spring lengths but not used them and youve introduced the spring constant, i would have expected the question to ask you to determine it / not have k in the solution.
Thanks. So values are okay? I will try to correct the ODE.

I used spring lengths to find spring constant. Why no k in the solution?

I found a solution online for general case: https://physics.stackexchange.com/qu...-on-an-incline
It is too hard for me, but it gets a + (k/m) x = gsin(theta)
So this is the correct equation I think.

I tried again.

Define x to be displacement in the direction from wall to block
F = ma = T - mgsin(theta)
extension = x-L = x - 1.2
so ma = mgsin(theta) - k(x-1.2) [is this step right?]
a = gsin(theta) - k(x-1.2)
so a + k(x-1.2) = gsin(theta)

so SHM because a + kx = 0 is homogeneous case of the DE above
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1 month ago
#10
(Original post by Minibus)
Thanks. So values are okay? I will try to correct the ODE.

I used spring lengths to find spring constant. Why no k in the solution?

I found a solution online for general case: https://physics.stackexchange.com/qu...-on-an-incline
It is too hard for me, but it gets a + (k/m) x = gsin(theta)
So this is the correct equation I think.

I tried again.

Define x to be displacement in the direction from wall to block
F = ma = T - mgsin(theta)
extension = x-L = x - 1.2
so ma = mgsin(theta) - k(x-1.2) [is this step right?]
a = gsin(theta) - k(x-1.2)
so a + k(x-1.2) = gsin(theta)

so SHM because a + kx = 0 is homogeneous case of the DE above
Im unsure about both your question and also your background/knowledge to solve it. If you're doing SHM at a-level you should have done quite a few examples which cover things like defining displacement, determining the spring constant given equilibrium/natural lengths, .... There is little direction in the question about what they want or how much working to get there. "Writing down" the answers involving k could get full marks the way its written, but I suspect not. Can you be a bit more specific about the question / your knowledge? You've started mentioning homogeneous ODEs, is this because of the stackexchanage post or have you covered them and their solution? As Ive said previously, you've got most of the elements in your previous posts and the answers are ~right, but the working/justification isn't clear/coherent, so it really depends how important this is for your understanding / the question.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#11
(Original post by mqb2766)
Im unsure about both your question and also your background/knowledge to solve it. If you're doing SHM at a-level you should have done quite a few examples which cover things like defining displacement, determining the spring constant given equilibrium/natural lengths, .... There is little direction in the question about what they want or how much working to get there. "Writing down" the answers involving k could get full marks the way its written, but I suspect not. Can you be a bit more specific about the question / your knowledge? You've started mentioning homogeneous ODEs, is this because of the stackexchanage post or have you covered them and their solution? As Ive said previously, you've got most of the elements in your previous posts and the answers are ~right, but the working/justification isn't clear/coherent, so it really depends how important this is for your understanding / the question.
The question is just a HW question on SHM. I have only done definition of SHM and the formulae for maximum speed, period, angular frequency, acceleration. Definition for SHM I use is a = -kx. (acceleration proportional to displacement from a fixed point, and are opposite in direction)

It is question meant for pre-university (A-Level), so I cannot solve differential equations. I just use the formulae directly

How can I improve clarity of working and justifications?
Last edited by Minibus; 1 month ago
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1 month ago
#12
(Original post by Minibus)
The question is just a HW question on SHM. I have only done definition of SHM and the formulae for maximum speed, period, angular frequency, acceleration. Definition for SHM I use is a = -kx. (acceleration proportional to displacement from a fixed point, and are opposite in direction)

It is question meant for pre-university (A-Level), so I cannot solve differential equations. I just use the formulae directly

How can I improve clarity of working and justifications?
If you're doing A levels and you're just meant to assume the formulae, then what you've done is sort of OK. If you want to structure it a bit better, do the following steps clearly - you've done some of them
1) Use force balance equilibrium to get spring constant k.
2) Formulate the homogeneous SHM ODE by considering perturtubations from the equilibrium position.
3) Get the general solution to the ODE - this emphasises why the a = +/-x gives totally different solutions so the sign is important.
4) That gives the amplitude and period
5) Differentiate which gives you the maximum velocity / speed.

I thought A levels considered the complementary function solution to SHM ODE, but can't say for definite. As I said, if you're just meant to quote formula, quite a bit of this is unnecssary.
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Thread starter 1 month ago
#13
(Original post by mqb2766)
If you're doing A levels and you're just meant to assume the formulae, then what you've done is sort of OK. If you want to structure it a bit better, do the following steps clearly - you've done some of them
1) Use force balance equilibrium to get spring constant k.
2) Formulate the homogeneous SHM ODE by considering perturtubations from the equilibrium position.
3) Get the general solution to the ODE - this emphasises why the a = +/-x gives totally different solutions so the sign is important.
4) That gives the amplitude and period
5) Differentiate which gives you the maximum velocity / speed.

I thought A levels considered the complementary function solution to SHM ODE, but can't say for definite. As I said, if you're just meant to quote formula, quite a bit of this is unnecssary.
Thanks!! That helps me a lot. For step 2, I was writing the ODE by using N2L first, but made sign error earlier. I think just considering the equilibrium and the pertubations is easier.

I rewrite my earlier solution with your steps:
1) Obtaining spring constant k
mgsin(theta) = kx
3 * g * sin(30) = k(1.2-0.8)
1.5g = 0.4k
k = 15g/4

2) It is at equilibrium, with mgsin(theta) and T acting in opposite directions and exactly cancelling out.
Then, the mass is moved down x cm. This creates an additional tension force (the additional force is equal to k(x_0+x) - kx_0 = kx), which will be proportional to displacement - it's opposite to displacement, so it causes SHM. The mgsin(theta) force only changes the equilibrium point, so my equation is
ma = -kx
where x is displacement from 120cm mark on the ramp.
w^2 = k/m = 15g/(4*3) = 5g/4

Is this right? Thanks for helping me
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1 month ago
#14
(Original post by Minibus)
Thanks!! That helps me a lot. For step 2, I was writing the ODE by using N2L first, but made sign error earlier. I think just considering the equilibrium and the pertubations is easier.

I rewrite my earlier solution with your steps:
1) Obtaining spring constant k
mgsin(theta) = kx
3 * g * sin(30) = k(1.2-0.8)
1.5g = 0.4k
k = 15g/4

2) It is at equilibrium, with mgsin(theta) and T acting in opposite directions and exactly cancelling out.
Then, the mass is moved down x cm. This creates an additional tension force (the additional force is equal to k(x_0+x) - kx_0 = kx), which will be proportional to displacement - it's opposite to displacement, so it causes SHM. The mgsin(theta) force only changes the equilibrium point, so my equation is
ma = -kx
where x is displacement from 120cm mark on the ramp.
w^2 = k/m = 15g/(4*3) = 5g/4

Is this right? Thanks for helping me
Agree with 1), but you normally define x (or e) to be the extension from the natural length. So by definition x=0 at the natural length and x=0.4 corresponds to what is the new equilibrium. It obviously gives the same result. Im assuming, without loss of generality, that the positive x direction is down the slope so the tension in the extended spring acts up the slope. This should be clearly marked on your diagram.

2) Sounds too hand wavy. Using the x definition given in 1), formulate (Newton 2) the acceleration of the mass as a function of the forces acting on it. Then transform "x" so that the resultant ODE is homogeneous and in standard SHM form. Without giving too much away, the transformation must be a translation of the origin from the spring natural length to the equilibrium point.

I was idly mulling this over earlier and if the question asked for max amplitude and velocity only, an energy argument would be much simpler. However, as they ask for the period (time), really you have to go down the SHM sinusoidal route.

Ill also repeat that this is a "Rolls Royce" solution for the problem. If they required all of this, they really should ask for the intermediate parts (determine k, show SHM ODE, derive solution) in the question. This is to try and ensure you understand SHM rather than being a "proper" solution to the question.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#15
(Original post by mqb2766)
Agree with 1), but you normally define x (or e) to be the extension from the natural length. So by definition x=0 at the natural length and x=0.4 corresponds to what is the new equilibrium. It obviously gives the same result. Im assuming, without loss of generality, that the positive x direction is down the slope so the tension in the extended spring acts up the slope. This should be clearly marked on your diagram.

2) Sounds too hand wavy. Using the x definition given in 1), formulate (Newton 2) the acceleration of the mass as a function of the forces acting on it. Then transform "x" so that the resultant ODE is homogeneous and in standard SHM form. Without giving too much away, the transformation must be a translation of the origin from the spring natural length to the equilibrium point.

I was idly mulling this over earlier and if the question asked for max amplitude and velocity only, an energy argument would be much simpler. However, as they ask for the period (time), really you have to go down the SHM sinusoidal route.

Ill also repeat that this is a "Rolls Royce" solution for the problem. If they required all of this, they really should ask for the intermediate parts (determine k, show SHM ODE, derive solution) in the question. This is to try and ensure you understand SHM rather than being a "proper" solution to the question.
Yes, at this point I am just trying to improve my understanding to make the SHM equation (differential equation)

I attempted 2 again
This is my diagram
https://imgur.com/a/AwmV3E0

I'm assuming if I measure x as positive is displacement down the slope, then a positive value for acceleration a is an acceleration down the slope too?

failed attempt:
ma = mgsin(theta) - T
T=-kx
ma = mgsin(theta) + kx
But this wrong

so it's ma = mgsin(theta) + T? is it always like this?
why is above one (failed attempt) wrong? is it because - sign already takes into consideration direction, so subtracting again negated it and made it incorrect? I am getting confused on the sign errors I am making @[email protected]

good attempt (I think):
continuing from ma = mgsin(theta) + T
so ma = mgsin(theta) - kx

from equilibrium mgsin(theta) = kx_0
so ma = kx_0 - kx
ma + k(x-x_0) = 0
so my'' + ky = 0, where y=x-x_0
so y'' + (k/m)y=0

so it is shm with equilibrium position at x = x_0
(for eq, I put y'' =0,
so y = 0,
so x-x_0 =0,
so x = x_0)

Thanks again for the help
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1 month ago
#16
(Original post by Minibus)
Yes, at this point I am just trying to improve my understanding to make the SHM equation (differential equation)

I attempted 2 again
Editing your response ...
> ma = mgsin(theta) + T
> ma = mgsin(theta) - kx

So this is assuming positive displacement, velocity, accleration is down the slope. So gravity is positive and when you have a positive extension, the tension = -kx is negative as its pulling you back up the slope. This is what you obtained. When the acceleration is 0, the kx = mgsin(theta), which is the equilibrium position. Again, this is a useful check.

However, this is close to the SHM ODE as you have acceleration proportional to the negative displacement, but there is an extra constant term, mgsin(theta).

> from equilibrium mgsin(theta) = kx_0
> so ma = kx_0 - kx
> ma + k(x-x_0) = 0
> so my'' + ky = 0, where y=x-x_0
> so y'' + (k/m)y=0

Yes. The non-homogeneous ODE has a constant forcing term (gravity). You can transform this into a homogeneous ODE by translating the displacement to the equilibrium point. As x_0 is constant y''=x'' (=a). This is now the standard, unforced SHM. You'd expect sinusoidal oscillations about the equilibrium point and parameter matching (guessing the solution) you say
omega^2 = k/m

If the ODE was
y'' - (k/m)y = 0
or equivalently
y'' = (k/m)y
You'd get exponential divergence rather than sinusoidal oscillations. So that +/- is very important.

Have you covered how to solve such ODEs? Complementary functions or ...
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#17
(Original post by mqb2766)
Editing your response ...
> ma = mgsin(theta) + T
> ma = mgsin(theta) - kx

So this is assuming positive displacement, velocity, accleration is down the slope. So gravity is positive and when you have a positive extension, the tension = -kx is negative as its pulling you back up the slope. This is what you obtained. When the acceleration is 0, the kx = mgsin(theta), which is the equilibrium position. Again, this is a useful check.

However, this is close to the SHM ODE as you have acceleration proportional to the negative displacement, but there is an extra constant term, mgsin(theta).

> from equilibrium mgsin(theta) = kx_0
> so ma = kx_0 - kx
> ma + k(x-x_0) = 0
> so my'' + ky = 0, where y=x-x_0
> so y'' + (k/m)y=0

Yes. The non-homogeneous ODE has a constant forcing term (gravity). You can transform this into a homogeneous ODE by translating the displacement to the equilibrium point. As x_0 is constant y''=x'' (=a). This is now the standard, unforced SHM. You'd expect sinusoidal oscillations about the equilibrium point and parameter matching (guessing the solution) you say
omega^2 = k/m

If the ODE was
y'' - (k/m)y = 0
or equivalently
y'' = (k/m)y
You'd get exponential divergence rather than sinusoidal oscillations. So that +/- is very important.

Have you covered how to solve such ODEs? Complementary functions or ...
Oh I get it now I think. First attempt wrong because it leads to always increasing displacement

I haven't done the solving of differential equations.

I think I am just meant to convert it to a = - constant * x, and write constant = (2pi/T)^2 to find period and everything else

Thank you for your time! I got it now
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1 month ago
#18
(Original post by Minibus)
Oh I get it now I think. First attempt wrong because it leads to always increasing displacement

I haven't done the solving of differential equations.

I think I am just meant to convert it to a = - constant * x, and write constant = (2pi/T)^2 to find period and everything else

Thank you for your time! I got it now
As a slight extension.

Without proving it, you can say that the solution is
y(t) = Acos(wt) + Bsin(wt)
Where the constants A and B are determined by the initial conditions (displacement and velocity) at time t=0 and w is omega. Zero initial velocity means B=0, so A=0.3 and
y(t) = 0.3cos(wt)
... Differentiating gives the velocity
y' = -0.3w*sin(wt)
...
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