Helical Spring Question: Determining Distance & Velocity in Vibrating Mass

[eddyb]

hi i am struggling with this question as i can't work out how to approach it. I am unsure which steps i need to take to resolve it and any help would be much appricated!

The question is

A mass suspended from a helical spring is set vibrating and it is found that 125 complete oscillations takes 75s. If the ampliltude of the motion is 145mm determine the distance from the equilibrium position when the velocity reaches 50% of its maximum velocity

really stuck cheers for any help in advance

 

[Päällikkö]

https://www.physicsforums.com/showthread.php?t=94379

Please show your own work. You must have at least some sort of general idea what equations might come in handy.

 

[eddyb]

i don't as the question i have been given is diffrent from all my notes usually you have mass or spring constnat (k) but i just really haven't got a clue where to start

 

[Päällikkö]

Well then I strongly suggest you hit the books.

Using the principle of conservation of energy, can you write an equation which links the amplitude, the speed at a given instant, and the displacement at that particular instant?

 

[eddyb]

i can't see how that helps thou i am given the amplitude motion is 145mm but i can't see how i am meant to link it. I've work out that the angular velocity is 10.47 so i used 10.47 0.0145 x cos 10.47 x 0.6 which is wACoswT which gives the velocity i got -0.052ms-1 i just catn get that intial step and work out what i need to find out

 

[Päällikkö]

145 mm = .145 m
Could you explain the equation you just used: Why does the cosine term have the period T in it?

T is actually not needed at all to solve this problem.PS.
To make your equations clearer could you use LaTeX, or just stick with symbols/letters (no numbers, please)?

 

[eddyb]

well i thought that finding the displacement of the spring would help in find the displacement at 50% of its velocity. So i used x(t) = Xm Cos wT It is used to find the elongation (x) of the spring.. The formlua i posted a minute ago is wrong i just went threw it. I know that the time is takes to complete one cycle is 0.6 seconds which is 1.6 hz. which i used to find the angular velocity... that's as far as i have got now i appricate the help

 

[Päällikkö]

Ok, the equation is still wrong. It should be:
x(t) = A \sin (\omega t)
(You could replace the sine with cosine.) The important part is that there is no T (well actually yes there is, "inside" w), and t is the variable.

You earlier took the derivative of this equation, so you ended up with the velocity:
v(t) = \omega A \cos (\omega t)

First of all, can you determine the maximum speed?
Now, can you determine when the mass has the speed half the maximum?

 

[eddyb]

ok so the equation for the velocity would be
v = 10.47 x .145 x cos (10.47 x 0.6)?

The amplitude i think is .145 and the angular velocity is 10.47 just can't work out what i am missing here. i think i can work out the speed at half by dividing the time by amplitude by 2?

 

[Päällikkö]

ok so the equation for the velocity would be
v = 10.47 x .145 x cos (10.47 x 0.6)?

Again, no.
v = \omega A \cos (\omega t)
where t is a variable, and has nothing to do with T, the period.

Thus v_{max} = \omega A
Can you figure out why this is true, and determine an instant the mass has maximum speed?

Now, can you determine an instant t so that v(t) = \frac{v_{max}}{2} ?

The amplitude i think is .145 and the angular velocity is 10.47 just can't work out what i am missing here. i think i can work out the speed at half by dividing the time by amplitude by 2?

No, you can't.

 

[eddyb]

ah i see now because T has no effect on the velocity at which it will fall an rise! thanks so i got Vmax = 10.47 x .145 = 1.51815ms-1 and at 50% of the maxiumum velocity t = 1.15815ms-1 / 2 = 0.759075ms-1. So to find the displacement at 50 % i need to use the velocity at 50%?

 

[Päällikkö]

at 50% of the maxiumum velocity t = 1.15815ms-1 / 2 = 0.759075ms-1.

How did you get this? It's wrong.

You get t from the equation:
v(t) = \frac{v_{max}}{2}

\omega A \cos (\omega t) = \frac{\omega A}{2}

 

[eddyb]

have i got the angular velocity and amplitude right? did i get the max velocity right then? sorry just can't quite how you find t or is t = wA/2 which would be 10.47*.145/2 ?

 

[Päällikkö]

have i got the angular velocity and amplitude right? did i get the max velocity right then?

They should all be correct. You however don't need the value of angular velocity to solve this problem.

sorry just can't quite how you find t or is t = wA/2 which would be 10.47*.145/2 ?

You need to solve the instant t the mass has the given speed of half the maximum:

v(t) = \frac{v_{max}}{2}

Now substitute for v(t) and v_max for the equation above:
\omega A \cos (\omega t) = \frac{\omega A}{2}

Solving for t:

t = \frac{\arccos(1/2)}{\omega}

Have you got any idea what to do next?

 

[eddyb]

i can't really read what that says does it say t = arccos(1/2)/w?

i know we need to solve t but i just can't work out the fiqures to put in this is what is making me strugle from the beginning i know w is 10.47 and that's it apart form what i explained earlier either i am being stupid or just missing something. I get the steps which you are kindly explaining but just can't work out where to get the fiqures from

 

[eddyb]

i got 4.29 for t=arccos(1/2)/w

 

[Päällikkö]

i can't really read what that says does it say t = arccos(1/2)/w?

It does, yes. You get it from the equation above it if you solve for t.

i know we need to solve t but i just can't work out the fiqures to put in this is what is making me strugle from the beginning i know w is 10.47 and that's it apart form what i explained earlier either i am being stupid or just missing something. I get the steps which you are kindly explaining but just can't work out where to get the fiqures from

Again, w is rather irrelevant, as its numerical value is not needed (it will cancel out).

By figures, do you mean like numbers? I don't usually substitute them for the symbols until the very end: this makes the solution easier to follow.Don't think about the steps we've taken.
We've solved an instant t the speed is half the maximum. We are asked for the displacement when the speed is half the maximum. Do you know what to do?

 

[eddyb]

so now we have t is t the half, or do we work out the half from t? then rearrange another equation to produce x which is the extrention of the spring?

 

[Päällikkö]

We already have an equation x(t).
The t we solved is the instant the speed is half the maximum. Thus, we substitute this t to the displacement equation x(t).

I strongly suggest revising simple harmonic motion.

Another way of solving the problem is through the principle of conservation of energy (which I suggested in the beginning). You might want to try if you can solve the problem that way.

 

[eddyb]

so all i need to do now is rearrange the equation again to find X which is the extention ?

 

[Päällikkö]

so all i need to do now is rearrange the equation again to find X which is the extention ?

Umm... I'm not quite sure what you mean. The asked displacement x:

x(t) = A sin(arccos(1/2)) = sqrt(3)/2 A = .126 m

 

[eddyb]

right thank you i know get it. doing another example now so the asked displacement is .126 m many thanks chap really do appricate your time and efforT!