What other information is needed to solve this equation involving b?

In summary: I do not understand what "clearing this for b" means. Specifically, what does "clearing" mean and what "this" is that needs to be "cleared".
  • #1
leomarrg
6
0
Homework Statement
Consider the movement of the mass and spring with friction
A * [e ^ - (b / 2m) t] * cos (sqrt (k / m - (b / 2m) ^ 2)) t + φ)
With A = 2m, k = 3N / m, m = 5Kg and φ = 0, we could change to the value of "b" from zero until the oscillations disappear for the first time and the function does not change to negative values. In this case, the friction now has a very noticeable effect: it lengthens the period of the oscillation so much that no oscillations will be seen (T -> infinity).
The theory indicates when that should happen. Clear for b from the equation and record the predicted value for b = .... (Kg / s) (at least two decimal places to the right of the point)
The question is complete, the exercise is clear for b, I don't get how to do it. The answer is 7.746
Relevant Equations
A * [e ^ - (b / 2m) t] * cos (sqrt (k / m - (b / 2m) ^ 2)) t + φ)
the equation is not equal not to anything (example: 3x = 6) and i don't know how to resolve [e ^ - (b / 2m) t].
 
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  • #2
leomarrg said:
Homework Statement:: Consider the movement of the mass and spring with friction
A * [e ^ - (b / 2m) t] * cos (sqrt (k / m - (b / 2m) ^ 2)) t + φ)
With A = 2m, k = 3N / m, m = 5Kg and φ = 0, we could change to the value of "b" from zero until the oscillations disappear for the first time and the function does not change to negative values. In this case, the friction now has a very noticeable effect: it lengthens the period of the oscillation so much that no oscillations will be seen (T -> infinity).
The theory indicates when that should happen. Clear for b from the equation and record the predicted value for b = ... (Kg / s) (at least two decimal places to the right of the point)
The question is complete, the exercise is clear for b, I don't get how to do it. The answer is 7.746
Relevant Equations:: A * [e ^ - (b / 2m) t] * cos (sqrt (k / m - (b / 2m) ^ 2)) t + φ)
leomarrg said:
the equation is not equal not to anything
Then it's not an equation.
 
  • #3
leomarrg said:
Homework Statement:: Consider the movement of the mass and spring with friction
A * [e ^ - (b / 2m) t] * cos (sqrt (k / m - (b / 2m) ^ 2)) t + φ)
With A = 2m, k = 3N / m, m = 5Kg and φ = 0, we could change to the value of "b" from zero until the oscillations disappear for the first time and the function does not change to negative values. In this case, the friction now has a very noticeable effect: it lengthens the period of the oscillation so much that no oscillations will be seen (T -> infinity).
The theory indicates when that should happen. Clear for b from the equation and record the predicted value for b = ... (Kg / s) (at least two decimal places to the right of the point)
The question is complete, the exercise is clear for b, I don't get how to do it. The answer is 7.746
Relevant Equations:: A * [e ^ - (b / 2m) t] * cos (sqrt (k / m - (b / 2m) ^ 2)) t + φ)

the equation is not equal not to anything (example: 3x = 6) and i don't know how to resolve [e ^ - (b / 2m) t].
Why does the friction change the period of oscillation? When the damped oscillator crosses zero, it is moving faster than at any other point since it last came instantaneously to rest. Yes, the period depends on bb but for a given bb it is constant.

What are you trying to find? If it is b, you need to know the position at some other time or the velocity at ##t=0##.
 
Last edited:
  • #4
You have ##x(t)=Ae^{-{\frac{b}{m}t}}\cos(\omega t).## What do you get when you substitute ##t=0## in this expression?
 
  • #5
kuruman said:
You have ##x(t)=Ae^{-{\frac{b}{m}t}}\cos(\omega t).## What do you get when you substitute ##t=0## in this expression?
I get 2
 
  • #6
leomarrg said:
I get 2

That is incorrect. When you substitute ##t=0##, you get
##x(0)=Ae^{-\frac{b}{m}*0}\cos(\omega*0)##. What does that simplify to?
 
  • #7
kuruman said:
That is incorrect. When you substitute ##t=0##, you get
##x(0)=Ae^{-\frac{b}{m}*0}\cos(\omega*0)##. What does that simplify to?
my bad, was using the data given in the exercise, A?
 
  • #8
leomarrg said:
my bad, was using the data given in the exercise, A?
Write the equation to which the equation in post #6 simplifies. ##A## is not an equation.
 
  • #9
kuruman said:
Write the equation to which the equation in post #6 simplifies. ##A## is not an equation.
x(0) = A?

e^−b/m * 0 = 1, and cos(w * 0) = 1
 
  • #10
leomarrg said:
x(0) = A?

e^−b/m * 0 = 1, and cos(w * 0) = 1
Yes. Now what? What is the question you want us to help you with?
 
  • #11
kuruman said:
Yes. Now what? What is the question you want us to help you with?
A * [e ^ - (b / 2m) t] * cos (sqrt (k / m - (b / 2m) ^ 2)) t + φ)
A = 2m, k = 3N / m, m = 5Kg and φ = 0
clearing this for b. The complete exercise is on the first question. If t = 0 then cos (sqrt (k / m - (b / 2m) ^ 2)) t is going to = 0. the answer is 7.746 but i don't know how to get there...
 
  • #12
I do not understand what "clearing this for b" means. Specifically, what does "clearing" mean and what "this" is that needs to be "cleared". Perhaps you can post an example in which something is "cleared" for b or whatever.
 
  • #13
kuruman said:
I do not understand what "clearing this for b" means. Specifically, what does "clearing" mean and what "this" is that needs to be "cleared". Perhaps you can post an example in which something is "cleared" for b or whatever.
I'm translating this question, it was given on a quiz it's original language is Spanish and i don't know if the correct term is clear for b, but essentially i want to know the value of b. That's what i mean by clear. Already have A, m, k, and φ
 
  • #14
The correct word is "solve" for b. OK, now that we have settled that, what do you know other than
##x(t)=Ae^{-\frac{b}{m}t}\cos(\omega t)## and the values for ##A##, ##m##, ##k## and ##\phi##?
 
Last edited:

1. What does "clearing for b" mean in a formula?

"Clearing for b" in a formula means isolating the variable b on one side of the equation by using algebraic operations.

2. Why is it important to clear for b in a formula?

Clearing for b allows us to solve for the specific value of the variable b and understand its relationship to the other variables in the formula.

3. How do you clear for b in a formula?

To clear for b, you must perform the inverse operation of what is being done to b. For example, if b is being multiplied by a number, divide both sides of the equation by that number to isolate b.

4. Can you clear for b in any type of formula?

Yes, you can clear for b in any type of formula as long as it contains the variable b and follows the rules of algebra.

5. Are there any common mistakes when clearing for b in a formula?

Some common mistakes when clearing for b include forgetting to perform the inverse operation on both sides of the equation, making a sign error, or forgetting to distribute a number or variable when needed.

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