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Oscillations concerning a pendulum

  1. May 10, 2015 #1
    1. The problem statement, all variables and given/known data

    A clock is regulated by a pendulum. The pendulum can be considered as a small weight connected to a rod of negligible mass. The period of oscillation of the pendulum can be adjusted by moving the weight up or down the rod. The angular frequency is given by ##\omega ^2 =\frac{g}{L}##, where ##L## is the distance between the centre of the small mass and the pivot point of the pendulum.

    The clock is placed on a space ship accelerating from the surface of the Earth with
    acceleration ##a=3g.## Calculate the period of vibration of the pendulum.


    2. Relevant equations

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


    3. The attempt at a solution

    The net acceleration is 2g.

    Therefore,

    ## T = 2 {\pi} \sqrt{\frac{(50 \times 10^{-2})} {(2g)}} ##

    ##T = 1.003544 = 1 s##

    Does this seem correct?
    Thank you!
     
  2. jcsd
  3. May 10, 2015 #2

    Suraj M

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    How did you get this?
     
  4. May 10, 2015 #3
    ##\sum F_y = ma = -mg + 3mg ##

    ##ma = 2mg ##

    ##a = 2g##

    Taking forces radially downwards as negative.
     
  5. May 10, 2015 #4

    Suraj M

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    Try to find the force acting on a body inside the shuttle rather than the total force acting on the space ship
    Hint: you should consider pseudo force on the pendulum by newtons 1st law
     
  6. May 10, 2015 #5
    Suraj, is this correct?

    ##\sum {F_y} = m(3g) = - mg + T ##

    ## T = m(4g) ##

    So,
    ## a = 4g ##

    Therefore,
    ## T = 0.71\ seconds##

    Thank you.
     
  7. May 10, 2015 #6
    Thank you !
     
  8. May 10, 2015 #7

    Suraj M

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    As you have taken forces downward as -ve ypu should be getting a as a -ve value, which you're not,
    What is the T you have introduced?
    also ##∑F_y ≠m(3g)##
    you got the right answer, but i doubt the method you have used..i do not mean to discourage you, but its better to know the right method, so that you can use it for another question.
     
  9. May 10, 2015 #8
    Oh, the T is the tension of the string, which is upwards. How would I go about this? It's true I would rather know the method so I can apply the method again in the future.
     
  10. May 10, 2015 #9

    Suraj M

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    Thats a good way to find out the total force acting on the bob, your T should be the resultant of 2 downward(-ve) forces, could you recognize those forces?
     
  11. May 10, 2015 #10
    sorry, since the rocket is accelerating upwards at 3g, does that mean the clock inside is also accelerating upwards at 3g?
     
  12. May 10, 2015 #11

    Suraj M

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    We'll it is, but when you're standing in a bus and the vehicle is accelerating forward at 5m/s² will you feel a forward force acting on you, or a backward force? apply your findings to the bob.
     
  13. May 10, 2015 #12
    So the bob will 'feel' a force of 3mg downwards. Adding this to its weight mg will result in the tension being 4mg upwards.
     
  14. May 10, 2015 #13

    Suraj M

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    yes the tension is 4mg upwards which is the one which balances out the downward force experienced by the bob, so you can get your new g' = a=-4g .. and you'll have your answer.
    Just remember about the concept of pseudoforce in cases like this, body feels a force equal and opposite to the direction of its (accelerated) motion.
    :smile:
     
  15. May 10, 2015 #14
    Thanks again Suraj.

    Kind regards and all the best!
     
  16. May 10, 2015 #15

    Suraj M

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    You're welcome, :smile:
    All the best to you too .
     
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