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Change in time period of pendulum

  1. Mar 4, 2016 #1
    1. The problem statement, all variables and given/known data

    Q The length of a simple pendulum executing SHM is increased by 21% .The percentage increase in the time period of the pendulum of increased length is
    a) 11%
    b) 21%
    c)42%
    d)10%

    2. Relevant equations

    ##T = 2\pi\sqrt{\frac{L}{g}}##

    3. The attempt at a solution

    ##T^2 = 4{\pi}^2\frac{L}{g}##

    By differentiating the above expression and dividing the resultant equation by the above equation, ##\frac{2dT}{T} = \frac{dL}{L}##

    ##\frac{2dT}{T}## x ##100 = \frac{dL}{L}## x ##100##

    ##\frac{dL}{L}## x##100 = 21##

    Therefore , ##\frac{dT}{T}## x ##100 = 10.5 ## . But this isn't correct .

    Please help me find the error in my approach .

    Many Thanks




     
    Last edited: Mar 5, 2016
  2. jcsd
  3. Mar 4, 2016 #2
    There is a really easy way to solve this question. If it increased by 21%, How can you put that in the equation.

    Lets just say that I am gaining profit of 5% every month. If my original balance is 100$, and I want to know my balance after one month. Well I can just calculate
    100 + 100 * 0.05
    Which can be 1.05(100)

    So you can do the same to the pendulum.

    About your equation above how did you derive that equation? can you post your steps?
     
  4. Mar 5, 2016 #3

    ehild

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    The approximation f(x+Δx)=f(x)+f '(x) Δx can be applied only when Δx/x << 1, which is not true now. Calculate the ratio between the two time periods from the original formula.
     
  5. Mar 5, 2016 #4
    Great ! :bow:

    Should that be Δx →0 ?
     
  6. Mar 5, 2016 #5

    ehild

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    You know Δx, it is a fixed quantity.
    In the limit, lim f(x+Δx)=f(x) when Δx→0.
     
  7. Mar 5, 2016 #6
    Sorry . I did not understand your reply .
     
  8. Mar 5, 2016 #7

    ehild

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    No, Δx is a fixed quantity. Ignore my second sentence.
     
  9. Mar 5, 2016 #8
    Ok . Thanks .
     
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