Effects of Removing a Spring on Oscillating Mass: Analysis and Amplitude Changes

[MeMoses]

Homework Statement

A mass is between 2 springs between 2 walls. The mass oscillates with amplitude d. The springs are at equilibrium when the mass is in the center. At the moment the mass is in the center a spring is removed. What is the resulting x(t) and the new amplitude? Note the k's are equal.
Edit: I forgot to add that at t=0 x(t)=d/2 when there is one spring.

Homework Equations

The Attempt at a Solution

So I solved for x(t) for both before and after removing the spring.
Before: x(t) = Asin(2wt) + Bcos(2wt)
after: x(t) = Asin(wt) + Bsin(wt)
I'm just assuming the new amplitude is twice what is was since when the spring is removed it has the energy of 2 springs, but now the resistance of one. Is there a way to show this using the x(t)'s I found? I feel like they should relate to the problem more.

 

[azizlwl]

1. With both in equilibrium, means not forces applied to the mass.
2. If the mass displaced with 2 springs, then with only one spring we have to take conservation of energy into consideration.

 

[MeMoses]

1. It is oscillating with amplitude d, we know forces are being applied.
2. I know we can use conservation of energy, but is there anyway using the x(t) that i derived?

 

[azizlwl]

Before: x(t) = Asin(2wt) + Bcos(2wt)
after: x(t) = Asin(wt) + Bsin(wt)
-------------------------------
The amplitude must be different for both cases.

 

[MeMoses]

for before I get just x(t) = Asin(2wt) since x(0)=0 but how do I show the amplitude difference with these equations? Edit: Actually scratch that, x(0) = d/2, I'll see where this gets me

 

[voko]

The As and Bs are different in the two equations. You need to express one set via the other. A and B in the initial equation must be expressible via d and the condition that at t = 0 the mass is going through the equilibrium point.

 

[MeMoses]

I can solve for B using x(0) = 0 but for A do i have to use the fact that Asin(wt) + Bcos(wt) should equal 2d at their max, so when wt = pi/4, and then solve for A?

 

[voko]

B (for the original equation) = 0, that's correct. So the original equation is x = A sin 2wt. What is A in terms of d?

 

[MeMoses]

Sorry. I fogot to include an important fact in the question. I just added it. Using that I get B = d/2, so how do I get A? What I was thinking in #7 was wrong, I'm guessing I just find a maximum of my function using derivatives, which won't work either, nevermind.

 

[voko]

Where does this additional condition come from? Is it part of the problem? What time is t = 0?

 

[MeMoses]

It is from the problem and t = 0 is when the spring disappears (also in the problem). I tried to only type the relevant stuff. I clearly didnt understand the problem fully

 

[voko]

This new condition contradicts the other statement. One spring is said to disappear when the mass is at the center. That means x(0) = 0. x = d/2 means the mass is half-way between the center and the maximum displacement.

 

[MeMoses]

Ok now that makes sense. It does not disappear when m is at the middle, but when x=d/2 at which t=0. I am really glad I read properly, but I'm still not sure how to find A or the new amplitude

 

[voko]

Does the problem specify in which direction the mass was moving when the spring separated?

 

[MeMoses]

Yes, to the right. How is that relevant?

 

[voko]

To get A, you have to simplify the equation

$$x = A \sin 2 \omega t - B \cos 2 \omega t$$

Divide it through by $\sqrt {A^2 + B^2}$, that will get you

$$x = \sqrt {A^2 + B^2}[ a\sin 2 \omega t - b \cos 2 \omega t ]$$

where $a \le 1, \ b \le 1$ so you could let $a = \cos \alpha, \ b = \sin \alpha$, and use trigonometry to transform the equation further.

 

[MeMoses]

Well don't I already know B=d/2 using x(0), so is there not an easier method?

 

[voko]

Using this method, you convert $x = A \sin 2 \omega t + B \cos 2 \omega t$ into $x = \sqrt {A^2 + B^2} \sin (2 \omega t + \alpha)$, and it is then obvious that $d = \sqrt {A^2 + B^2}$.

 

[MeMoses]

I honestly have no idea how or why/how you did that or how you decided that it needed to be done, nor do i see how d is obvious.

 

[voko]

Re how I did that, go the previous page and how another look.

Re why d is obvious: look at $x = \sqrt {A^2 + B^2} \sin (2 \omega t + \alpha)$ - what is the maximum value of it? Then recall the definition of amplitude.

That explains WHY I did that.

 

[MeMoses]

OK that makes more sense, just a few more question. You said divide it through by sqrt(A^2 + B^2). If dividing by that, how does the sqrt end up being multiplied by the sin and cos? Also where do a and b come from? why must they be <=1? and why can we just let them = cos(alpha) and such? I'm just a little lost with this method

 

[MeMoses]

also if d = sqrt(A**2 + B**2) isn't that d for the amplitude of just one spring, which isn't known? A isn't known as well so it doesn't seem to get me anywhere.

 

[voko]

Little a and b are big A and B divided by the common factor; admittedly, "divide through" was a bit inaccurate an expression, more properly it should be "extract common factor". Can you see now why they should satisfy the inequalities?

As for the trigonometric representation of a and b, imagine a unit circle centered at the origin of the XY plane. Now take x = a and y = b. They will be within the unit circle due to the inequalities they satisfy. So they will correspond to ONE point at the circle, and that point also have a corresponding angle. And a and b are then expressed via that angle.

 

[voko]

also if d = sqrt(A**2 + B**2) isn't that d for the amplitude of just one spring, which isn't known? A isn't known as well so it doesn't seem to get me anywhere.

The problem states that d was for two springs. You have obtained B = d/2, so from this equation you can determine A.

 

[MeMoses]

So I thought $x = \sqrt {A^2 + B^2} \sin (2 \omega t + \alpha)$ was equal to x(t) for one spring. Your saying that solving amplitude for it gives me d (from 2 springs) but the equation uses A and B from x(t) with just one spring? If its equivalent to x(t) for 1 spring we don't know that value of amplitude, if its equivalent to x(t) for 2 springs, we don't know A or B

 

[voko]

All the time till now I have been discussing the ORIGINAL equation with two springs. We need to determine its constants A and B in order to obtain the velocity at time t = 0 and thus have initial conditions for subsequent motion.

 

[MeMoses]

So then A for the original equation is sqrt(d**2 - d**2/4) and then the velocity from the 2 springs at t=0 is the same as the velocity of one spring at t=0 so set them equal and then A=2sqrt(d**2-d**2/4) for one spring. The same d = sqrt(A**2+B**2) should apply here as well if I am not mistaken.

 

[voko]

The two-spring A is indeed $\sqrt {d^2 - d^2/4} = d \frac {\sqrt 3} 2$.

I do not follow the rest of your argument, however. You have not yet computed the velocity at t = 0, but you seem to be jumping at conclusions about subsequent motion.

 

[MeMoses]

At t=0 (the moment the spring disappears) v(0) using the equation from 2 springs will equal v(0) using the equation for 1 spring, which gives w*d*sqrt(3)=w*A => A = d*sqrt(3). That was the last unknown for the 1 spring equation. Using the same method for the 2 spring system then the amplitude = sqrt(A**2 + B**2) and everything has been answered.

 

[voko]

So what is the amplitude for the 1-spring system?

 

[MeMoses]

sqrt(3*d**2 + d/2*sqrt(3))

 

[voko]

I don't understand how you got sqrt(3) inside sqrt().

By the way, I think you made a mistake early on when you got x = A sin 2wt + B cos 2wt. How did you get the 2 there?

 

[MeMoses]

damn it should be sqrt(2) not 2 in there, and before that mistake I had A=d*sqrt(3) and B=d/2, A^2 = 3*d^2 and B^2 d/2*sqrt(3)

 

[MeMoses]

And fixing that mistake I get A in the 1 spring to equal d/2*sqrt(6). I then get amplitude = sqrt(6*d**2 / 4 + d/2*sqrt(3))

 

[voko]

So the final result is?