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SHM Problem

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle performs S.H.M. with amplitude 25 cm and period 3 s. The minimum time required for it to move between two points 12.5 cm on either side of the mean position is ?

    2. Relevant equations
    ##y##=##a## ##sin####w####t##

    3. The attempt at a solution
    Solution is
    ##y##=a sin ##w## t

    6.25=25 sin##\frac{2π}{3}##t

    t=0.25

    I don't understand why would we take y=6.25?
    I don't understand what " the minimum time required for it to move between two points 12.5 cm on either side of the mean position is " it means?
    I take it as total time required to go from +12.5 to mean position and then from mean position to -12.5 cm.

    it will take ##\frac{T}{4}## time to reach +a from mean position i.e to reach 25 cm from mean psition
    then it will take T/8 time to reach mean position from +12.5 cm . And it will take another T/8 time to reach -12.5cm from mean position hence total time T/8 + T/8 = 2T/8 = 0.25 T
    Right?
    SHM.png
     
    Last edited: Apr 22, 2016
  2. jcsd
  3. Apr 22, 2016 #2

    blue_leaf77

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    Just want to clarify, by 12.5 cm, is it the distance between the two points, or the distance between each point from the mean position?
     
  4. Apr 22, 2016 #3
    I don't know. I don't have any extra information . What do you think?
     
  5. Apr 22, 2016 #4

    blue_leaf77

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    The reason I asked you was because I don't know either. What about the solution you provided, is this part:
    all from the one who gave you the problem? The solution for t=0.25 does not satisfy that equation.
     
  6. Apr 22, 2016 #5
    I was practicing questions based on SHM. I found the following test
    http://career.webindia123.com/career/entrance/ques.asp?sub=physics&groupID=9&course_name=&page=2
    Look at question number 7.
    Then I started searching for it's solution and I found this
    http://www.raoiit.com/MH-CET-2015-Sol/Medical-UG%20MH-CET%202015%20Question%20Paper%20version-11.pdf [Broken]
    Jump to question number 5
    Then I looked for it's solution and finally I got
    http://www.raoiit.com/MH-CET-2015-Sol/MH-CET-%202015-Answer%20Key%20and%20Solutions%20version-11.pdf [Broken]
    QN5.png
    See question number 5.
     
    Last edited by a moderator: May 7, 2017
  7. Apr 22, 2016 #6

    blue_leaf77

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    For some reason I couldn't access the last two links. Nevertheless I tried to work out the problem the way it should be done; it turns out to be the correct answer, which is different from the first solution t=0.25 s.
    So, what about your own attempt? Have you checked if it's correct?
     
  8. Apr 22, 2016 #7
    You tell me. Am I correct ? that's what my OP is all about.
     
  9. Apr 22, 2016 #8

    blue_leaf77

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    At the bottom end of the first link in post #5, there is a "submit" button you can use to check your answer, you don't need to answer the rest of the question to see your result of problem no. 7. Your answer is 0.25x3 s = 0.75 s which is not among the choices.
    That's not true for non-linear functions, such as sinusoids.

    Imagine your particle starts from the mean position at t=0. Then the displacement equation looks like ##y(t) = 25 \sin \omega t##. Find the time at which the particle reaches ##y=12.5## for the first time.
     
  10. Apr 22, 2016 #9
    I did not multiply it by 3.
     
  11. Apr 22, 2016 #10

    blue_leaf77

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    Then what does T in
    stand for?
     
  12. Apr 22, 2016 #11
    Sorry! o:)
     
  13. Apr 22, 2016 #12
    When should I use π=3.14 and when to use π=180 degrees?
    I think when we are measuring angles we should use π=180 degrees otherwise π=3.14
    so in ##w##=##\frac{2π}{T}## π is 180 degrees, right?
    But I have seen somewhere frequency of vibrations=f=##\frac{w}{2π}## and then omega value is given to be 4
    f=2/3.14
    we can see here π=3.14 in spite of the fact that π is denoting angle.
     
  14. Apr 22, 2016 #13

    blue_leaf77

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    You use 3.14 when the input to the sine function of your calculator is in radians, and use 180 degrees when the input is in degrees. Both values denote an angle, but in different unit.
     
  15. Apr 23, 2016 #14
    But I have noticed that when π is with sin , cos functions we take it to be 180 degrees and when it isn't accompanied by any such functions it is taken to be 3.14. Why?
     
  16. Apr 23, 2016 #15

    cnh1995

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    In calculus, all the angles are in radians. So π is always 3.14 in calculus. For example, when we write y=sin(2πft) and dy/dt=(2πf)cos(2πft), π is 3.14. But when we express some sinusoidal quantity in this way, y=sin(ωt+∅), ∅ can be written in degrees for our understanding. For example, we write
    y=sin(ωt) and y1=sin(ωt+π) to show that the signals have 180° (or π radian) phase difference. Here, π represents phase difference and can be read as 180°(for our understanding). But when it comes to differentiation or integration, π is always 3.14.
     
  17. Apr 24, 2016 #16
    ##y(t)##=##25## sin ##w## ##t##

    12.5 =##25## sin ##\frac{2π}{3}## ##t##

    =##25## sin ##\frac{360}{3}## ##t##

    =##25## sin 120 ##t##

    =##25## 0.86 ##t##

    ##\frac{12.5}{25×0.86}## =##t##

    0.58=##t##
     
  18. Apr 24, 2016 #17

    blue_leaf77

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    No no no, you totally messed up there. The variable ##t## is included in the argument for the sine function: ##\sin (\omega t)##.
     
  19. Apr 24, 2016 #18
    12.5= ##25## sin (120 t)
    0.5 = sin (120 t)
    (120 t) = 30
    t=30/120
    t=0.25
     
  20. Apr 24, 2016 #19

    blue_leaf77

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    Yes. Now you just need to find the time needed for the particle to travel from +12.5 to mean position and then from mean position to -12.5 cm.
     
  21. Apr 24, 2016 #20
    I don't have to include time taken to travel from mean position to 12.5 i.e time I just calculated.
     
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