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Ratio of masses of bob

  1. Jun 22, 2014 #1
    1. The problem statement, all variables and given/known data
    i have attached the sample ans here.
    hand written photo is my working.

    2. Relevant equations



    3. The attempt at a solution
    since the question doesnt same string . so i assume it's a same spring. so i would get t is directly propotional to sqrt root (m). is my concept wrong?
     

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  3. Jun 22, 2014 #2

    Simon Bridge

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    The question states that the string is elastic - so what effect would different masses of bob have on the string?
    Note: the question asks you to explain the effect - your working is maths only, an explanation would have words in it.
     
  4. Jun 22, 2014 #3
    post deleted~
     
    Last edited: Jun 22, 2014
  5. Jun 22, 2014 #4

    Simon Bridge

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    I dunno - I cannot follow it.
    That is why I'm asking about your reasoning.
     
  6. Jun 22, 2014 #5
    hi, i am just asking the maths part.the spring is elastic, so the extension of spring would be different for both cases. why cant I use the above working ?since this is a spring-mass system. so T=2pi(m/k) 1/2... taking 2pi(k)^-0.5 as constant. T is directly propotional to m
     
  7. Jun 22, 2014 #6

    Simon Bridge

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    The x in F=kx is the extension ... that is the change in length of the spring.
    The pendulum string has an unstretched length as well.
    Did you take that into account?

    How did you deal with the string changing length during the swing?
     
  8. Jun 22, 2014 #7
    in other words, i let (m/k)=(l/g) since mg=kl. so i change the ratio of (l/g) into
    (m/k)... so i get T is directly propotional to m
     
  9. Jun 22, 2014 #8

    ehild

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    You assumed the bob vibrating vertically (first picture), while the problem speaks about a pendulum, the bob moving along an arc of circle about the equilibrium point (second picture). As it is an elastic string, the bob can move in a quite complicated way. If it is initially in rest, and you pull a little bit vertically, it performs vertical SHM with ω=√(k/m). That is not pendulum motion.
    If you kick the bob horizontally, it moves like a pendulum in the vertical plane. In case the deviation from vertical is very small, the length of the string does not change appreciably. For that motion, ω=√(g/l), the same as that of a "normal" pendulum. You need to use that formula, keeping in mind, that the length of the pendulum is not proportional to m. The string stretches because of the weigth of the bob, and the change of the length is proportional to the mass.


    ehild
     

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  10. Jun 22, 2014 #9
    i can understand it finally. thanks
     
  11. Jun 22, 2014 #10

    ehild

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    You are welcome

    ehild
     
  12. Jun 22, 2014 #11

    Simon Bridge

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    It's a common wrinkle in teaching labs where we deliberately give students a slightly stretchy string and get them to investigate the effect of changing the mass on the period - though, the size of the bob (as weights are added) is usually a bigger effect.

    Students are startled to discover a mass dependence - it's character building ;)

    Related - pendulum on a spring.
    https://www.physicsforums.com/showthread.php?t=538269

    The only niggle I have at the back of my mind is that this is the sort of question were the person setting it may be making some unspoken simplifications... kinda relying on the limited knowledge of the class to narrow the possible options.
     
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