Physics Challenge :)

[gopunk]

this is from a hw that i just did... i know the answer... trust me. i'm not tricking you guys into doing my hw for me! the numbers are changed in case my instructor sees this and thinks i'm cheating.



A platform of mass 0.64 kg is supported on four springs. A chunk of modeling clay of mass 0.12 kg is held above the table and dropped so that it hits the table with a speed of v= 0.5 m/s.

The clay sticks to the table so that the the table and clay oscillate up and down together. Finally, the table comes to rest 4 cm below its original position.

With what amplitude does the platform oscillate immediately after the clay hits the platform?




and btw, if any of you want to post your own questions, feel free... preferably stuff with waves and oscillation (that's what i need practice on)

 

[bizmark]

hmmm, it's been a while, and I don't have my physics books here (took it a few years ago), but let me see if I can get it.

The one thing that I can't do is account for four springs. If (and this might be correct) the four springs act essentially as one spring, then the following holds. If not, well, do something to account for that (might be *really* complicated), and it'll work.

So first we wanna find the spring constant. We do this by using the formula for the force that a spring provides, F=-kx, where x is displacement and k is the spring constant. So initially the spring is providing (for simplicity let's not worry about the signs in this, we're only working in one dimension and we know what's up and what's down, and what should be positive and what should be negative.... it'll all work out) (9.8m/s^2)*.64kg=6.272N of upward force. At the end, it's providing (9.8m/s^2)*.76kg=7.448N of upward force. So 6.272=k*x_0 and 7.448=k*(x_0+.04m), cancelling the k*x_0 and solving for k, we get that k=29.4N/m.

Now we know the work equation for a spring, W=-1/2*kx^2. The basic formula for Work is W=K_f-K_i, if we want to find the amplitude of the spring then we want to find out its displacement when it's extended to its maximum, i.e. when K_f=0. Our K_i was (1/2)*.12kg*(.5m/s)^2=.015 units of kinetic energy, and at the end it's got 0, so the negatives cancel, we've got .5*k*x^2=.015, so x=.93914m.

Is that right? Did I follow the right steps? As I said, I don't know about the 4 springs / 1 spring thing, but other than that.... I'll keep checking back. It's been a long time, so I'm probably rusty, but still I used to be pretty good Hopefully I don't look like an idiot right now

Would you mind explaining the difference that comes into play with multiple springs? What about multiple springs in series vs. in parallel, i.e. one long string of springs hooked together versus a row of springs connecting two blocks? How do the spring constants add together in these different situations, and how does that affect the period/etc.?

 

[Capn]

Ok, here's my solution:

The new weight added to the platform (which was at equilibrium) displaces it .04m

So:

(.12kg)(9.8m/s²)=k*x=k*(0.04m)

Therefore k = 29.4 N/M

Now we want to find the unloaded spring length. Which will be now known as D.

m*g=k*D
(.64kg+.12kg)*(9.8m/s²)=(29.4N/m)*(D)

Thus, D = .2533 m

We're going to approach this from an energy standpoint, or E1 = E2. In the real world this doesn't really work cause your system is damped, but for early conditions and assuming a low damping coefficient it's pretty darn good.

E1 = kinetic E + Spring potential E + gravitational potential E
E2 (at top of oscillation) = Spring Potential E + Gravitational potential E

I'm using the final resting spot as my zero for gravitational potential, also this is where 'x' is going to be measured from

E1 =(1/2)(.12kg)(0.5m/s)²+ (1/2)(29.4N/m)(.2533m-.04m)²+(.64kg+.12kg)(9.8m/s²)(.04m)
E1=.9817 J

E2=(1/2)(29.4N/m)(.2533m-x)²+(.64kg+.12kg)(9.8m/s²)(x)

Setting E1 = E2, then solving

x=+/-.0512m

So the amplitude of the oscillation is 5.12cm

For brownie points the natural (undamped) frequency of the system is ~39Hz

 

[SagaLore]

<< this is from a hw that i just did... i know the answer... trust me. i'm not tricking you guys into doing my hw for me! the numbers are changed in case my instructor sees this and thinks i'm cheating.



A platform of mass 0.64 kg is supported on four springs. A chunk of modeling clay of mass 0.12 kg is held above the table and dropped so that it hits the table with a speed of v= 0.5 m/s.

The clay sticks to the table so that the the table and clay oscillate up and down together. Finally, the table comes to rest 4 cm below its original position.

With what amplitude does the platform oscillate immediately after the clay hits the platform?

and btw, if any of you want to post your own questions, feel free... preferably stuff with waves and oscillation (that's what i need practice on) >>



Okay, so we have to set the variables.

m = 120 (.12 x 1000)
v = .5
r = .004 (4 x .001)

Have to find A.

Then we begin reverse factoring:

n = (m + q) / i
n = (120 + 11) / 2.2
n = 59.54

p = (v - 1) / (n + 1) - 1
t = pn (pn + v) - v

t = ((v-1) / (n + 1) - 1) ((m + q) / i) (((v-1) / (n + 1) - 1) ((m + q) / i) + v) - v

And don't forget we need the co-efficient of r in relation to n,

r(n) = (r1 + r2 + r3) / r(-1)

n = (n1 + n2 + n3) / -1(n)

Which gives us a formula of:

t = ((v-1) / (((((m + q) / i)'1 + ((m + q) / i)'2 + ((m + q) / i)'3 / -1((m + q) / i)) + 1) - 1) ((m + q) / i) (((v-1) / (((((m + q) / i)'2 + ((m + q) / i)'3 / -1((m + q) / i)) / -1((m + q) / i)) + 1) - 1) ((m + q) / i) + v) - v

Now let's begin substituting:

t = ((.5-1) / (((((120 + q) / i)'1 + ((120 + q) / i)'2 + ((120 + q) / i)'3 / -1((120 + q) / i)) + 1) - 1) ((120 + q) / i) (((.5-1) / (((((120 + q) / i)'2 + ((120 + q) / i)'3 / -1((120 + q) / i)) / -1((120 + q) / i)) + 1) - 1) ((120 + q) / i) + .5) - .5

Now, that given the constant of q is quantum fluctuations in the oscillations (a reciprocal affect of wave theory within the magnetosphere of the earth), and that i represents the intertia force (solved by the acceleration of gravity), we now determine that

t = 4

So, then the time capture (t) is multiplied by r, we end up with .016 as the change of final position. The reciprocal of this solution is equal to the amplitude of the oscillation after the immediate impact:

A = 62.5 sine/pi

That was pretty easy. Next question?

 

[gopunk]

Now we know the work equation for a spring, W=-1/2*kx^2. The basic formula for Work is W=K_f-K_i, if we want to find the amplitude of the spring then we want to find out its displacement when it's extended to its maximum, i.e. when K_f=0. Our K_i was (1/2)*.12kg*(.5m/s)^2=.015 units of kinetic energy, and at the end it's got 0, so the negatives cancel, we've got .5*k*x^2=.015, so x=.93914m.

Is that right?

i don't think so.. you're right that the 4 springs part doesn't matter... but the system's kinetic energy is different from 0.015 because it includes the mass of both the platform and the clay, and the velocity is different (conservation of momentum).

 

[gopunk]

Now we want to find the unloaded spring length. Which will be now known as D.

m*g=k*D
(.64kg+.12kg)*(9.8m/s²)=(29.4N/m)*(D)

Thus, D = .2533 m

this is my first physics class in a few years, so sorry if this is dumb, but doesn't that find the displacement if you were to drop the platform with the clay at the same time?

i got a different answer, ignoring gravity because i took the eq. point as zero. i'll post it in a few hours...

 

[Capn]

Gopunk, I worked in absolute energy regarding the spring, because I tend to get my wires crossed if I work too much in relative states.

Anyhow, if the platform were taken off the spring, from the final resting point the spring would have a length of .2533m, this value is only then used for the spring potential calculations. Not sure if that helped ya any, maybe if you rephrase your question I could understand what you're looking for.

 

[bizmark]

<< i don't think so.. you're right that the 4 springs part doesn't matter... but the system's kinetic energy is different from 0.015 because it includes the mass of both the platform and the clay, and the velocity is different (conservation of momentum). >>



But conservation of energy has to hold too. At the start, there is no kinetic energy in the system with the springs -- only potential energy in the spring, and the potential energy from the gravity pulling on the mass of the platform. These two energies cancel each other out, since there's no movement initially. The only kinetic energy, at the start of the problem, comes from the ball of clay. It has a definite KE and it transfers all of it to the clay-platform system, since this is a lossless collision (it stuck to the platform -- okay, so in real life there would be a loss of energy because of deformation of the clay ball, but that would be impossible to account for in a basic physics class )

So initially, the velocity is 0.5m/s. After hitting the platform, the velocity for the ball-platform combo is approx. 0.199m/s. The KE stays the same at .015.

 

[PowerEngineer]

It's been way too long since I took physics, but here's my take...

I agree that the spring constant is 29.4 N/m based on the mass of the modelling clay and the lower resting position. I also agree with wbwither's focus on conservation of energy in the table/clay system. The KE before collision is 0.5*0.12*(0.5)^.5 or 0.015 N (as wbwither says). Of course this KE is all eventually dissipated by heating the spring before the table/clay system comes to rest at 4 cm lower than the table started. So let's start at this so-called rest state and think about what would happen if we could suddenly reinject that 0.015 N back into the system. The energy has to be absorbed by the spring, so (assuming no losses on the first swing) the potential energy increase in the spring (0.5*K*X^2) must increase by 0.015 N in order to bring KE to zero (i.e. downward velocity to zero). X = SQRT(2*0.015/K) = 0.03194 m or 3.194 cm. That's +/-3.194 cm around the -4.0 cm resting point.

Okay...this is a pretty interesting result to me because it seems to show that the kinetic energy and the spring constant are the only two factors that determine the magnitude of the oscillations. Cool!

Edited to correct for stupid math error!!! (wbwither, I think you made one too -- ours should be the same)

 

[bizmark]

Powerengineer's take on it makes sense to me.

He looks at the kinetic energy dissapated and then puts it back in. My reasoning was to look at when the kinetic energy had been wholly transformed into the spring's potential energy, i.e. when the platform "hit bottom" so to speak. Then the spring had done that exact amount of work on the system.

I dunno, does my way not make sense to anyone else? I know my result seems pretty big (nearly a meter), so maybe there's a factor that I missed or something? Also I didn't take gravity into account in the second part. Maybe that was my problem?

Funny how we're all getting different answers and I really don't see anything wrong with anybody else's reasoning

 

[Capn]

I don't think Powerengineer's method is correct and here's why.

He's going off of some sort of conservation of energy method, but he's doing it incorrectly. At the initial state you do have .015 J of kinetic energy, but you also have gravitational potential and spring potential depending on how you set up your system. You can't just take one portion of the energy, and inject it back into the system at a different state. I'm not saying my solution is absolutely correct cause I can make some simple errors from time to time, but Powerengineer's method definitely is not correct.

wbwither looking at your solution I can't figure out how you calculated x=.93914m

using this equation:
.5*k*x^2=.015

I get x=3.2cm, which is the number Powerengineer found, and by the same reasoning above your solution is incorrect as well. Just by inspection we can understand this solution has to be incorrect, the magnitude of oscillation has to be greater than 4cm. Your solution would be correct if the kinetic energy were given at the point of equilibrium of the spring-mass system, however we are given kinetic energy at a point 4cm above this new equilibrium point.

I'm going to solve this from a dynamic systems point of view and see what I get

x(0) = .04m
x'(0) = -.0789m/s (conservation of momentum)

Forgetting about gravity and using the equilibrium point of our new system, then setting up free body diagram, there's one force of:

k*x

Setting up our second order ODE using D'Alembert's principle and summing forces yields:

k*x - m*x" = 0

The general solution of this ODE is:

x = A*cos(omega*t + phi)

Where A is magnitude, omega is angular frequency of the system (in radians/s) and phi is the phase

omega = sqrt(k/m) = 6.22 (I have no idea how I calc'ed 39 hz before, hmn oops)

at t =0, remember IC's

x(0) = .04m
x'(0) = -.0789m/s (conservation of momentum)

so...

x'=-omega*A*sin(omega*t + phi)

-.0789 = -(6.22)*A*sin(phi)

.04 = A*cos(phi)

Two equations, two unknowns....

tan(phi) = .3173
phi = .3073

A = .04/cos(phi) = .042

So, my updated answer would be 4.2cm I guess, must've done something wrong in that other example. Is this the number you got Gopunk?

 

[gopunk]

A = .04/cos(phi) = .042
So, my updated answer would be 4.2cm I guess, must've done something wrong in that other example. Is this the number you got Gopunk?

i've very sorry i haven't had time to look at everybody's stuff today

yes, that is what i got... i put it in my profile's icq number so i wouldnt' forget. i typed out all my work and pm'ed it to myself, thus wasting 15 minutes because apparently self-sent pm's do not show up

here is how i did it:

i got the spring constant, like you guys all did. 29.4 N.

using conservation of momentum, i determined the system's speed the instant the clay hits:

0.64 kg + 0.12 kg = 0.76 kg
(0.12 kg)(0.5 m/s) = 0.06 kg m/s
(0.06 kg m/s) / 0.76 kg = 0.07895 m/s

from there i used an energy viewpoint. i know that total energy at the instant clay hits is PE + KE, and that when it goes down the amplitude, E = PE.

Initially:
KE = (0.5)(0.76 kg)(0.07895 m/s)² = 0.002368 J
PE = (0.5)(29.4 N)(0.04 m)² = 0.02352 J
E = 0.002368 J + 0.02352 J = 0.02589 J

At amplitude:

PE = 0.02589 J = (0.5)(29.4 N)(x²)
0.001761 m² = x²
x = 0.04197 m

 

[bizmark]

<< I get x=3.2cm >>



Hmm funny I get that too now... oops.. calculator error anyone?

Ooh, just thought of the fact that conservation of energy only holds for perfectly elastic collisons. My bad

Hmmm well I can't understand your work Capn (I couldn't understand SagaLore's work either) but it seems that you know what you're talking about. And if you and Gopunk (whose work I could understand ) got the same answer using two different ways, then.. I believe it.

 

[MereMortal]

<< Ooh, just thought of the fact that conservation of energy only holds for perfectly elastic collisons. My bad >>



No. Conservation of energy holds for any collision. Conservation of kinetic energy, however, only holds for perfectly elastic collisons.

 

[PowerEngineer]

Yes, I messed that one up a bit I should have taken the potential energy into account and needed to use conservation of momentum rather than mistakenly assuming conservation of energy at collision time. Now you know why I decided to go electrical...my gut is more closely aligned with Faraday than it is with Newton.