What is the Speed of an Object in Simple Harmonic Motion?

[FaNgS]

Homework Statement

The spring constant of the spring is 23N/m and the mass of the object is 6kg. The spring is unstretched and the surface is frictionless. A constant force of 16N is applied horizontally to the object causing the spring to stretch.
Find the speed of the object after it has moved a distance of 0.32m.

Answer range: 0.470372 - 3.23253

Homework Equations

F=-mxw^2
v=w(A^2 - x^2)^(1/2)

The Attempt at a Solution

F=-w^2mx
where k=mw^2
16=kx , x=16/23 m ---Amplitude A

v=w ( (16/23)^2 - (0.32)^2 )^(1/2)

v= 1.20935 m/s (wrong)

 

[denverdoc]

not sure this is SHM problem at all. Can you try this from work-energy theorum?

 

[FaNgS]

it is actually a SHM problem...we were given this question while we were studying SHM and we were asked to solve it

but i understand...i also thought the same thing but my intructor said use SHM theories and equations for this problem

 

[denverdoc]

well if the block ever gets released, it will certainly become one, no doubt. Or I suppose it could be a mass attached to the springmass system falling over the side of the table and at some point the spring will recoil. Hmmm.

 

[denverdoc]

Best I can come up with is then to look at this from first principles,

as in mx"+kx=Fapp,
the homogeneous soln is periodic motion, and then you need to solve for the particular case, which will also be periodic but with some change in coefficients.

 

[FaNgS]

no you got the question wrong...the spring is attached a wall and in turn attached to a block mass 6kg..it will not fall over it will remain on a frictionless surface the whole time

i'm really sorry for not making the question clear

 

[FaNgS]

the problem is finding the amplitude of the oscillation for the object

i don't see any relation with the given information to calculate the amplitude A so i can use the equation v=w (A^2 - x^2)^(1/2) unless there's another equation I'm missing??

 

[denverdoc]

No my bad, if its finding the amplitude, you can easily figure out where things change direction, ie when the net acceleration is zero. I thought the question asked for velocity

ma=-kx+16 so kx=16/m. which I think you had posted to begin with. I'm trying to do too many things at once here.

 

[FaNgS]

so now if i calculate x ( x=16/k*m) will x be the amplitude??

 

[FaNgS]

if x from the equation u gave is the amplitude A and x is 0.32m and if i subsitute it into the equation v=w(A^2 - x^2)^(1/2) it won't work because the difference A^2 - x^2 is negative and u can't have a square root of a negative number.

:( *sob*

 

[lalbatros]

I don't understand your "relevant equations".
You should at least define your symbols and explain you ideas.

If F is the force,
if x is the displacement (from the zero-force location),
if k is the spring constant,
then, I think we have by definition: F=kx.

Therefore, you should be able to find the initial position of the mass.
The rest of the solution is a simple application of the energy conservation.
You know the intitial position,
therefore you know the intitial potential energy,
and the initial kinetic energy is zero.
Now, if the mass moves to another position it is a simple think to calculate the potential and kinetic energies and the speed.

(I got v=1.14636 m/s, is that right?)

 

[FaNgS]

lalbatros if possible can u show ur steps ?

 

[lalbatros]

FaNgS,

Well, from F=kx, you get x=F/k=16/23 m.
From there on, there are many different ways to solve the problem.
You need to work that out by yourself, otherwise it is not useful.
I give a sketch.

Energy method
Remember that the total energy of the system is the kinetic energy plus the potential energy.

Etotal = mv²/2 + kx²/2

and this is constant.
The statement of the problem is not very clear, but we should guess that the spring is streched to an equilibrium position where F=16N and where the mass is kept immobile. Therefore, in the initial position:

x=16/23 m and v=0 m/s

later, x = 16/23-0.32 m and since the total energy remains the same, you can calculate the speed. Job done.

Harmonic motion method
You also know that the motion will be given by:

x(t) = 16/23 Cos(Wt)

where I chose a cosine to fit the position for t=0,
and where W²=k/m=23/6 s^-2 is the pulsation of this motion.
By derivating it is easy to find the speed for any time t: v = -16/23 W Sin(Wt) .
It is also easy to find the time when the mass comes into position x = 16/23-0.32 m.
Combining these two reasoning also solve your problem.
Note: it will not be necessary to calculate t explicitely, since all you need to calculate is: Sin(Wt)=(1-Cos(Wt)²)^0.5

Remark
The results obtained byt the first and the second method will be the same.
This does not happen by chance!
Actually, from the energy conservation written in the first method, it is possible to derive the harmonic solution used in the second method.

Please take the time to work out the solution.

Postsciptum
As a further exercice, you could check that from the harmonic solution the total energy is indeed conserved, in general.