Simple Harmonic Motion Problem 4

[thunderhadron]

Hi friends the problem is -

https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/s480x480/155412_2656530589803_1383873256_n.jpg

Attempt -

As per the problem states,

When the compound system will oscillate in its natural frequency, The frequency of the oscillation will be, √[k/(m + M)]

When m will fall through the height h, it'll gain √2gh speed. For some time let say it u.

It will impact with the cage. Using conservation of linear momentum in the vertical direction,

linear momentum before = linear momentum after collision.

so,

m.u = (m + M)v

so v = m.u/(M + m)

and this position will be the mean position of the oscillations,

Here the cage + particle system would be at it maximum speed which would be,

v = m.u/(M + m)

The maximum speed at the mean position is, v = A.ω

So, A = v/ω

Putting the value of v and ω

I am getting the result like this-

A = {√[k / (M + m)]} . m√(2gh)/(M + m)

Which is not the correct answer as per the question states.

The correct answer of this problem is option (A) as per the question.

Now I have doubt in the question. Is it correct ? We can not Conserve the K. E. of the system because the collision is not perfectly elastic.

Please friends help me in solving this Problem.

Thank you all in advance.

 

[ehild]

Hi friends the problem is -

https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/s480x480/155412_2656530589803_1383873256_n.jpg

Attempt -

As per the problem states,

When the compound system will oscillate in its natural frequency, The frequency of the oscillation will be, √[k/(m + M)]

When m will fall through the height h, it'll gain √2gh speed. For some time let say it u.

It will impact with the cage. Using conservation of linear momentum in the vertical direction,

linear momentum before = linear momentum after collision.

so,

m.u = (m + M)v

so v = m.u/(M + m)

and this position will be the mean position of the oscillations,

Here the cage + particle system would be at it maximum speed which would be,

v = m.u/(M + m)

The maximum speed at the mean position is, v = A.ω

So, A = v/ω

Putting the value of v and ω

I am getting the result like this-

A = {√[k / (M + m)]} . m√(2gh)/(M + m)

.

Check your result, you multiplied by ω instead of dividing with it. Otherwise your derivation is correct, and the result given are wrong. The amplitude should depend on both masses.

ehild

 

[TSny]

linear momentum before = linear momentum after collision.
so,
m.u = (m + M)v
so v = m.u/(M + m)

and this position will be the mean position of the oscillations

Is this true? Where would the system hang at rest with both masses M + m?

Here the cage + particle system would be at it maximum speed

Need to reconsider this if the mean position of the oscillation is not the collision point.

Now maybe I'm not seeing something. I don't get anything close to any of the answers given.

 

[AlephZero]

There is a sneaky way to do questions like this. You don't have find the right answer yourself. You just have to select which of the given answers is right

Look at the four answers and ask yourself
- Which of the fornulas gives a length? The formulas that don't give a length must be wrong.
- Of the formulas that are left, which ones don't contain some variable that will obviously affect the answer? (Don't do the math, just use common sense). Those formulas are also wrong.

 

[ehild]

Is this true? Where would the system hang at rest with both masses M + m?



Need to reconsider this if the mean position of the oscillation is not the collision point.

Now maybe I'm not seeing something. I don't get anything close to any of the answers given.

You are right, the initial position of the cage ΔL₀=Mg/k is not same as the mean position of oscillation ΔL=(M+m)g/k.

Taking downward positive, the SHM y=Asin(ωt+θ) starts with y₀=-mg/k and v₀=um/(m+M).
Anyway, the given answers can not be correct.

ehild

 

[thunderhadron]

Look at the four answers and ask yourself
- Which of the fornulas gives a length? The formulas that don't give a length must be wrong.
- Of the formulas that are left, which ones don't contain some variable that will obviously affect the answer? (Don't do the math, just use common sense). Those formulas are also wrong.

Well the first statement is good. I am getting answer with its help. But maths is not allowing this. Now I am thinking that, I have taken the mean position in wrong way. So all the calculation is wrong. But the formulas I used, I think I am correct in that.

If I leave this technique also(i.e using the mean position and the maximum speed during SHM), applying energy conservation is also not providing the correct answer.

 

[haruspex]

So, A = v/ω
A = {√[k / (M + m)]} . m√(2gh)/(M + m)

You appear to have multiplied by ω instead of dividing by it.

 

[thunderhadron]

You appear to have multiplied by ω instead of dividing by it.

haruspex:

The general equation of SHM is, x = A sinωt.

SO the velocity of the particle would be,

v = Aω cosωt

This velocity will be maximum when cosωt will be maximum and ±1

so V_max = ± Aω

hence, A = v/ω

 

[haruspex]

haruspex:

The general equation of SHM is, x = A sinωt.

SO the velocity of the particle would be,

v = Aω cosωt

This velocity will be maximum when cosωt will be maximum and ±1

so V_max = ± Aω

hence, A = v/ω

I have no problem with any of that. It was the next line that I quoted, where you seem to have done A = ωv instead.

 

[SammyS]

Hi friends the problem is -

https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/s480x480/155412_2656530589803_1383873256_n.jpg

Attempt -

As per the problem states,

When the compound system will oscillate in its natural frequency, The frequency of the oscillation will be, √[k/(m + M)]

When m will fall through the height h, it'll gain √2gh speed. For some time let say it u.

It will impact with the cage. Using conservation of linear momentum in the vertical direction,

linear momentum before = linear momentum after collision.

so,

m.u = (m + M)v

so v = m.u/(M + m)
...

It seems to me that there is another factor to be considered. That is, during the time in which the body (of mass m) is in free fall, the cage begins an upward acceleration (the beginning of SHM motion, with ω' = √(k/M) ) since the cage-spring system will not be in equilibrium.

 

[TSny]

It seems to me that there is another factor to be considered. That is, during the time in which the body (of mass m) is in free fall, the cage begins an upward acceleration (the beginning of SHM motion, with ω' = √(k/M) ) since the cage-spring system will not be in equilibrium.

Oh, interesting! So, we are to assume the mass m was initially attached to the cage before it fell. Back to the drawing board for me.

 

[TSny]

Thanks, SammyS. I believe it does indeed work out to be answer (A). [EDIT: Dang. Found an error. Not getting a simple answer.]

 

[haruspex]

That is, during the time in which the body (of mass m) is in free fall, the cage begins an upward acceleration (the beginning of SHM motion, with ω' = √(k/M) ) since the cage-spring system will not be in equilibrium.

I see nothing in the problem statement that implies the cage-spring was not in equilibrium initially. Indeed, if you don't assume that there's not enough info to solve the problem.

EDIT: Ok, I see what you're suggesting - that the mass m was initially on some support within the cage, then fell from it. If that's correct,m the problem is very poorly worded, but I admit that with the more obvious reading I get none of the offered amplitudes. (I get the rebarbative √{(2mg/k)(mh/(M+m) + (M+m)g/k)}.)

 

[SammyS]

Thanks, SammyS. I believe it does indeed work out to be answer (A). [EDIT: Dang. Found an error. Not getting a simple answer.]

Well, I haven't worked it all the way through myself, but I suspect the answer should depend on both M and m , so answer (A) looks suspect. (Other answers ruled out per AlephZero)

 

[SammyS]

I see nothing in the problem statement that implies the cage-spring was not in equilibrium initially. Indeed, if you don't assume that there's not enough info to solve the problem.

I agree.

I am referring to the time during which the body, of mass m, is in free fall. I assume that before the body is in free fall, the cage-body-spring system is in equilibrium.

 

[TSny]

I see nothing in the problem statement that implies the cage-spring was not in equilibrium initially.

I think SammyS is interpreting the problem as saying that the mass m was initially supported inside the cage at a height h above the bottom of the cage (maybe hanging from a string attached to the top of the cage). Everything is initially at rest in equilibrium when suddenly the string supporting m breaks.

 

[haruspex]

I think SammyS is interpreting the problem as saying that the mass m was initially supported inside the cage at a height h above the bottom of the cage (maybe hanging from a string attached to the top of the cage). Everything is initially at rest in equilibrium when suddenly the string supporting m breaks.

Yes, just realized that. I edited my earlier post.
But I don't think it's going to lead to a solution. We have to compute the KE remaining after impact. To get that, we need the collision speed, and that appears to involve solving a mixture of functions equation.

 

[TSny]

(I get the rebarbative √{(2mg/k)(mh/(M+m) + (M+m)g/k)}.)

For the record, I get almost the same result for the original interpretation. Only difference is that I have m instead of (M+m) in the last term.

 

[thunderhadron]

I have no problem with any of that. It was the next line that I quoted, where you seem to have done A = ωv instead.

OMG. Sorry Really I did mistake out there

 

[thunderhadron]

I agree.

I am referring to the time during which the body, of mass m, is in free fall. I assume that before the body is in free fall, the cage-body-spring system is in equilibrium.

Well sorry to interrupt but my motive is to solve the problem. And I think any assumption couldn't work there SammyS.

 

[TSny]

OK, we noticed that we weren't getting any of the answers listed. So, SammyS thought there might be a different interpretation of the setup.

Going back to the original interpretation, you need to see that after m sticks to the cage and the SHM begins, the mean position of the oscillations is not the location of the collision. The oscillations will take place around the new equilibrium position where the system would hang at rest with total mass m+M.

 

[thunderhadron]

Initially at t = 0, when cage was at rest mass 'm' had gravitational potential energy of magnitude, "mgh"

When it'll hit with the cage it'll impart (m√2gh) / (M + m) = v speed to the mass + cage system.

According to the law of conservation of Mechanical energy,

Loss in the P.E. of the mass(at the time of impact) = Gain in K.E. of (mass + cage)

And

Loss in K.E. of (mass + cage) at maximum elongation of spring = Gain in P.E. of spring.

So, mgh (I) = 1/2 . (m + m). v² (II) = 1/2. k.x² (III)



Equating (I) and (II),

mgh = 1/2 . k. x²

so x = [2mgh/k]^1/2

It is the correct answer according to the question and according to the person who made the question but he forgot that when the (cage + mass) system will go down then there will be loss in its gravitational P.E. too, Hence the equation will become like,

Loss in the P.E. of the mass(at the time of impact) = Gain in K.E. of (mass + cage)

And

Loss in K.E. of (mass + cage) at maximum elongation of spring = Gain in P.E. of spring + loss in P.E. of (mass + cage) System.

So, mgh = 1/2 . (m + m). v² = 1/2. k.x² - (M + m). g.x



Now it is not going to give the correct answer.

 

[thunderhadron]

Going back to the original interpretation, you need to see that after m sticks to the cage and the SHM begins, the mean position of the oscillations is not the location of the collision. The oscillations will take place around the new equilibrium position where the system would hang at rest with total mass m+M.

Tsny,
Can we interpret the position where the particle strikes with the cage. It is not the mean position. Mean position would be at (m + M)g/k. But Where the collision occur which position will be this. Is this any arbitrary location of the system?

 

[TSny]

Yes, the mean position of the oscillations will be at the point where the spring is stretched a distance (m+M)g/k. The collision takes place when the cage is just hanging at rest initially. How much is the spring stretched at the collision position?

 

[SammyS]

Initially at t = 0, when cage was at rest mass 'm' had gravitational potential energy of magnitude, "mgh"
...

My "assumption", which you refer to below, is that at the very beginning of the problem, immediately before the mass begins to fall, at that moment the system is at rest and therefore in equilibrium.

You seem to be assuming that also.

Well sorry to interrupt but my motive is to solve the problem. And I think any assumption couldn't work there SammyS.

 

[ehild]

Tsny,
Can we interpret the position where the particle strikes with the cage. It is not the mean position. Mean position would be at (m + M)g/k. But Where the collision occur which position will be this. Is this any arbitrary location of the system?

We can assume that the cage was in rest initially, and it did not change the position during the collision.

Initially the spring was stretched by Mg/k. It is above the new position by mg/k.
The cage + mass system will oscillate about the mean position after the collision. y=Asin(ωt+θ), and v=Aωcos(ωt+θ). Taking the zero of time at the instant of collision, the initial position of the shm is y(0)=Asin(θ)=mg/k, and the velocity is downward, with magnitude m/(m+M)u: v=Aωcos(θ)=-m/(m+M)u. or
Acos(θ)=-m/(m+M)u/ω. ω=√(k/(m+M)).

Squaring the equations in bold and adding them, you get A², which is different from answer A. That answer is wrong, as it does not contain the mass M. Imagine, will the amplitude be the same if you drop a bean onto bathroom scales or pharmacy scales?

ehild

 

[haruspex]

Initially at t = 0, when cage was at rest mass 'm' had gravitational potential energy of magnitude, "mgh"

When it'll hit with the cage it'll impart (m√2gh) / (M + m) = v speed to the mass + cage system.

According to the law of conservation of Mechanical energy,

Loss in the P.E. of the mass(at the time of impact) = Gain in K.E. of (mass + cage)

No, you can't use conservation of energy here. There's an inelastic collision. You correctly used conservation of momentum to find v. h cannot play any other role.
Under this interpretation (that the spring was initially extended only by the mass M, the mass m not being attached to it in any way), the equilibrium position will be mg/k below the cage starting position. So at this position the KE will be (M+m)v²/2+(M+m)g(mg/k). If the amplitude is A, the cage+mass, at its lowest point, will have lost a further (M+m)Ag in gravitational PE, and all will have been converted to PE in the spring. A is therefore is given by
(M+m)v²/2+(M+m)g(A+mg/k) = kA²/2
m²gh/(M+m) +(M+m)mg²/k = kA²/2 - (M+m)gA
Yuk.

 

[TSny]

A neat thing about a mass oscillating on a vertical spring is that if you let x measure displacement from the equilibrium position (rather than the total stretch of the spring) then you can forget about gravity in the oscillations. The mass will oscillate about the equilibrium position where the net force is -kx and you can take the potential energy of the system to be kx²/2. Then energy conservation relates the position just after the collision (where x = -mg/k) to the amplitude position (where x = A):

(m+M)v²/2 + k(mg/k)²/2 = kA²/2

where v = √(2gh) m/(m+M) is the speed just after the collision.

Then you can solve for A. (Same result as echild.)

 

[SammyS]

We can assume that the cage was in rest initially, and it did not change the position during the collision.

Initially the spring was stretched by Mg/k. It is above the new position by mg/k.
The cage + mass system will oscillate about the mean position after the collision. y=Asin(ωt+θ), and v=Aωcos(ωt+θ). Taking the zero of time at the instant of collision, the initial position of the shm is y(0)=Asin(θ)=mg/k, and the velocity is downward, with magnitude m/(m+M)u: v=Aωcos(θ)=-m/(m+M)u. or
Acos(θ)=-m/(m+M)u/ω. ω=√(k/(m+M)).

Squaring the equations in bold and adding them, you get A², which is different from answer A. That answer is wrong, as it does not contain the mass M. Imagine, will the amplitude be the same if you drop a bean onto bathroom scales or pharmacy scales?

ehild

It seems to me to be more likely that ehild's interpretation of the problem is the way it's author intended it to be interpreted. (Rather than what I was suggesting.)

 

[ehild]

It seems to me to be more likely that ehild's interpretation of the problem is the way it's author intended it to be interpreted. (Rather than what I was suggesting.)

And I am inclined to believe that the author intended to be interpreted the problem as thunderhadron did, (he forgot the new mean position) and also forgot to write in a factor of m/(M+m) somewhere.

ehild

 

[haruspex]

A neat thing about a mass oscillating on a vertical spring is that if you let x measure displacement from the equilibrium position (rather than the total stretch of the spring)

which I believe I did

then you can forget about gravity in the oscillations.

For the frequency, yes, but whether you can ignore it for amplitude depends on initial conditions. Often, the initial condition is a known displacement from equilibrium, and in that case the amplitude will be that displacement. But here, we know the velocity at eq.pos. As it descends, the spring has to take up both that KE and some extra lost PE. So the stronger the gravity the greater the amplitude.

 

[ehild]

There would be nonzero amplitude even in the case when the mas m had no speed, but just added to M.

ehild

 

[haruspex]

There would be nonzero amplitude even in the case when the mas m had no speed, but just added to M.

ehild

Yes, but what was that in response to?

 

[ehild]

Yes, but what was that in response to?

it was in response to you, supporting the statement that the amplitude depends on g. something like (gm/k) sqrt[1+2hk/(g(M+m))]. But your equation in post #27 does not look correct.

ehild

 

[haruspex]

But your equation in post #27 does not look correct.
ehild

It's horrid alright, but I can't find a flaw. Do you see one?