What Is the Maximum Extension of the Spring in Simple Harmonic Motion?

[pb23me]

Homework Statement

An ideal spring has a spring constant k=25N/M and is suspended vertically. A 1kg object is attatched to the unstrechted spring and released. It then performs ocillations.
What is the magnitude of the acceleration of the body when the extension of the spring is at maximum?
What is the maximum extension of the spring?

Homework Equations

F=-kx=ma
a=-A\omega²sin(\omegat)
x=sin(\omegat)
v=A\omegacos(\omegat)=0 ...at max extension

The Attempt at a Solution

i plugged -A\omega²sin(\omegat) into a for -kx=ma and got \omega=5 and I am completely stuck at this point.

 

[Maybe_Memorie]

You're forgetting the SHM defining equation; a = -(w^2)x

What will x be when the extension is maximum?

What is the time taken to go from min to max extension?

 

[pb23me]

ok that's cool but that equation doesn't help me much... i already know that w=5 and i have know idea of the acceleration or what x is at max or the time. If i knew either of those three i could solve it

 

[Maybe_Memorie]

Yes but you can very easily find all of those.

When it goes from one point of extremety to the other, what fraction of a complete period has it completed?

What is the period of the motion?

 

[pb23me]

would that be half a period? not sure

 

[pb23me]

period equals 1.26

 

[pb23me]

so x would equal .315

 

[pb23me]

w=5rad/sec
T=1.26s
t=.315s at max extension
still don't know how to find a value for A?
maybe I am missing an equation relating A to T?

 

[Maybe_Memorie]

Sorry, just realized getting T was pretty pointless..

Anyway, when a spring is fully extended (or compressed), what's it's distance from the equilibrium position?

 

[pb23me]

its the maximum distance away idk what numerical value that would be it would depend on the length of the spring...

 

[pb23me]

i just wrote out an equation f=ma=kx-mg=0 so x = .392m?

 

[pb23me]

however if i should have used this equation earlier f=ma=kx-mg when calculating w then i have the wrong value for w...w should have been equal to 3.9rad/sec

 

[tiny-tim]

hi pb23me!

i just wrote out an equation f=ma=kx-mg=0 so x = .392m?

mx'' = -kx -mg, so (x - mg/k)'' = -(k/m)(x - mg/k)

 

[pb23me]

first of all what does x" mean? secondly how did you establish that x"=-kx-mg?
the equation I am using is F=-kx. i was thinking that when the spring is at rest with the mass on it then F=ma=-kx+mg=0 i don't know if that force equation is correct however. i have know idea how you derived those equations

 

[tiny-tim]

x'' is acceleration (d²x/dt²).

I used F = ma (ie F = mx''), together with F = -kx.

This is for any x, not just the rest position.

(oh, I was out by a factor of m, i'll edit that in now )

 

[pb23me]

ok i get the first part mx"=-kx-mg it seems like this would be the case as the spring is moving down because gravity and the force of spring are in the same direction. however the second part (x - mg/k)'' = -(k/m)(x - mg/k) I am not sure what your doing here...it looks like your taking the second derivative of (x-mg/k) and getting -(k/m)(x - mg/k) ? keep in mind I am in a trig based physics class. Also I am trying to solve for x

 

[tiny-tim]

uhh? just expand it (and remember (constant)'' = 0)

(and now I'm going to bed :zzz:)

 

[pb23me]

Hi tiny tim, I am just confused about what force equation to use. F=-kx=ma? or F=-mg-kx=ma? I don't know if your supposed to include mg in your calculations with harmonic motion or not.

 

[tiny-tim]

yes, the F in F = ma is always the total force, so you must include everything …

in this case, that means both the mg and the kx

 

[pb23me]

awsome, so i have two more questions now
I used f=-ka=ma earlier to solve for w and got the correct answer by plugging in -Aw²sin(wt) for a and Asin(wt) for x. I verified this result with another equation i just came across T=(m/k)^1/2(2pi) then plugged T into w=2pi/T and came out with the same answer for w. This is just a little confusing if this equation is not the "complete" equation. Secondly on the equation you showed me earlier with F=-kx-mg=ma I was wondering why both of the signs are in the same direction?

 

[tiny-tim]

I used f=-ka=ma earlier to solve for w and got the correct answer by plugging in -Aw²sin(wt) for a and Asin(wt) for x. I verified this result with another equation i just came across T=(m/k)^1/2(2pi) then plugged T into w=2pi/T and came out with the same answer for w. This is just a little confusing if this equation is not the "complete" equation.

ah, that looks like the solution for a horizontal spring

Secondly on the equation you showed me earlier with F=-kx-mg=ma I was wondering why both of the signs are in the same direction?

oh, i decided to measure x and a upward …

but if you want to measure them downward instead, that's also fine!

 

[pb23me]

ok well F=-kx=ma = -kAsin(wt)=m[-Aw²sin(wt)]=-k=mw²
w=5rad/sec and then T=(m/k)^1/2(2pi) ...T=1.26sec so 2pi/T=w w=5rad/sec
thats a bit confusing idk. Can you just take an equation that's true such as F=-kx=ma -force of spring on an object- plug in other equations and come up with correct values?
Also F=-kx-mg=ma ... -k[Asin(wt)-mg=m[Aw²sin(wt)] so -mg=Asin(wt)[k-mw²] so i established earlier that sin(wt) = 1,-1 from v=Awcos(wt)=0 therefore wt = pi/2 or 3pi/2...so -mg=A[k-mw²] then A=.196 or -.196
a=-Aw²sin(wt) a=4.9 or -4.9m/sec²

 

[tiny-tim]

i'm getting very confused

can we go back to the original question, and check what we have to find …

What is the magnitude of the acceleration of the body when the extension of the spring is at maximum?
What is the maximum extension of the spring?

if that's all, then why do we need the simple harmonic motion, ω and so on (it is shm … but does that matter here)?

can't we just get the maximum extension from conservation of energy, and the acceleration at maximum extension from dividing the force there by the mass?

 

[pb23me]

this equation mgyi+1/2mv²i+1/2kxi=mgyf+1/2mv²f+1/2kx_f?? i really don't know what values to use for y and x are they the same? Or is this equation even right?

 

[tiny-tim]

this equation mgyi+1/2mv²i+1/2kxi=mgyf+1/2mv²f+1/2kx_f?? i really don't know what values to use for y and x are they the same? Or is this equation even right?

yes, x is the same as y, they're both height

that's the "KE + PE = constant" equation …

mgy and 1/2 kx² (you missed out the "squared" ) are the PE, and 1/2 mv² is the KE …

your v_i and your x_i = y_i are all zero, and for the maximum extension, you also put v_f = 0

sooo … the maximum extension is … ? ​

 

[pb23me]

ok but how is both vi=0 and vf=0? unless your were talking about the spring from top to bottom. but then you would be calling the top or the bottom the zero point for x or y?? i worked it out like that and got .784m

 

[tiny-tim]

ok but how is both vi=0 and vf=0?

that's what the original question said …

A 1kg object is attatched to the unstrechted spring and released. It then performs ocillations.

the object is attached (at zero speed) to the spring when its extension is also zero

(and of course the maximum extension must also have v = 0)

… unless your were talking about the spring from top to bottom. but then you would be calling the top or the bottom the zero point for x or y??

the top is the zero point for the spring … the question calls it the position in which the spring is unstretched, ie at x = 0 in all the usual spring equations

 

[pb23me]

so A would be .392m?

 

[tiny-tim]

so A would be .392m?

let's see … from conservation of energy, when v = 0 again (at x = 2A),

we have mgx = 1/2 kx², so x = 2g/k, ie A = g/k = .392 … yes!

 

[pb23me]

awsome i wonder why i got half of that the other way i worked it... F=-kx-mg=ma
-k[Asin(wt)]-mg=m[-Aw²sin(wt)] where w=5rad/sec from T=(m/k)^1/2(2pi) then w=2pi/T=5rad/sec so-mg=m[-Aw²sin(wt)]+k[Asin(wt)]
spring is at max extension so sin(wt)=1 I concluded this because v=Awcos(wt)=0 at max extension so it follows that wt = pi/2 or 3pi/2 therefore sin(wt) = 1 or -1 ...then
oh nevermind i see now it ends up being -g=A[-w²+k] or -g=A[w²-k]
dang! that almost worked lol