[[S.H.M]] Rearranging equations Confusion, Please HELP?

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Homework Help Overview

The discussion revolves around the rearrangement of equations related to simple harmonic motion (S.H.M.), specifically focusing on the calculation of the spring constant for a baby bouncer based on the mass and time period provided.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster expresses confusion regarding the rearrangement of the formula for the time period of oscillation and the appearance of the term 4π² in the equation for the spring constant. Participants discuss squaring the equation and applying algebraic rules to clarify the transformation of terms.

Discussion Status

Participants are actively engaging with the mathematical relationships involved, with some providing hints and guidance without revealing direct answers. There is a collaborative effort to clarify the steps necessary to arrive at the correct rearrangement of the formula.

Contextual Notes

The original poster mentions a lack of access to their tutor for assistance, which adds to their need for clarification on the mathematical principles at play.

Jack_Legacy
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[[S.H.M]] Rearranging equations Confusion, Please HELP?

The question that’s got me stuck
Question: A baby of mass = 9kg bounces with a time period of 1.2s in a baby bouncer. What is the spring constant k for the bouncer? Now I know I must rearrange the formula to find k
The formula I/the book used

T = 2π √m ÷ k

I do that and end up with (m × 2π) ÷ T² = k, My book on the other hand rearranges the formula as been (m × 4π²) ÷ T² = k.

Where does the 4π² come from I do not see how 2π changes too 4π², am I missing a rule? :confused:
Please help clarify how to get the 4π². I would ask my tutor but he is currently off sick.
My book also does this on the previous question where I am asked to find the value of l in the equation: T = 2π √l ÷ g my book rearranges the equation to (T² × g) ÷ 4π² = k :confused:
Again where does the 4π² come from in the original equation.

Thanks you for your help
Jack :confused:
 
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[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]

Square this equation.

Daniel.
 
maybe a dumb question but, what exactly do you mean? if i square on both sides don't i only cancel the square root and end up with T² on the other side e.g. I'd end up with
T² = 2π (m ÷ k) is this correct
 
Jack_Legacy said:
maybe a dumb question but, what exactly do you mean? if i square on both sides don't i only cancel the square root and end up with T² on the other side e.g. I'd end up with
T² = 2π (m ÷ k) is this correct

Do u agree that [tex](ab)^{2}=a^{2}b^{2}[/tex] ?

If so,apply it...


Daniel.
 
still don't quite understand i do agree with that but am quite sure how to apply it, i feel that you won't tell me the exact answer but could you guide me a little closer?
 
But of course i won't tell u the answer.The point is to help u find it by yourself.
HINT:
[tex](2\pi)^{2} =...??[/tex],knowing that:
[tex](ab)^{2}=a^{2}b^{2}[/tex]

Daniel.
 
Okay i think i have it: so if
T = 2π √m ÷ k and we want to find k then:
2π ÷ T = √m ÷ k then we square both sides
(2π ÷ T)² = (√m ÷ k)² which simplified is 4π² ÷ T² = m ÷ k
we then bring over the m and end up with m × 4π² ÷ T² = k
is that correct?
 
If it's something like that
[tex]k=\frac{4\pi^{2}m}{T^{2}}[/tex]

,then it's okay.


Daniel.
 

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