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Finding the constants in a Sine equation.

  1. Aug 9, 2008 #1
    1. The problem statement, all variables and given/known data
    I have been puzzling over this question for hours now.

    The centre of a wall clock is 180 cm above the floor. The hand of the clock that indicates the seconds is 20 cm long. The height, h cm above the floor, of the tip of the second hand, t seconds after midday, is given by an equation of the form: h(t)= a sin n (t+b) + c
    where a, b, c and n are positive real constants.

    2. Relevant equations

    I'm required to find what the value of the four constants is.

    3. The attempt at a solution

    I know that a=20 and c=180, but I'm not sure about n and c.
    Would it be correct if I assume that the period is 60 seconds. Therefore, to find n:

    60=2[tex]\pi[/tex][tex]/[/tex]n
    n=[tex]\pi[/tex][tex]/[/tex]30

    And to find c, is it right if I take t=0 and substitute all the values I've got of a, c, and n into the equation h(t)= a sin n (t+b) + c ?

    Also, what's the significant of the word midday? Am I supposed to take it as the start of the period?

    I feel really stupid for not understanding.
    I would really appreciate any help given, I'm at my wits end. Please?
     
  2. jcsd
  3. Aug 9, 2008 #2

    Defennder

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    The time given "midday" is telling you for which time during the day the equation accurately describes the clock, although there really isn't reason to believe that the equation would be different if the hour and minute hands point elsewhere. Just interpret it to mean that the equation tells you that h(0) gives you the distance from the tip of the second-hand to the floor when it is pointing at digit 12.

    You don't have to assume the period is 60s, though it probably is. Just imagine, suppose you have the second hand pointing at the digits 3,6,9,12. What is the distance from the tip to the floor in each case? Use this info, together with the equation to find the unknown values.
     
  4. Aug 9, 2008 #3
    Thanks! :smile:
     
  5. Aug 9, 2008 #4

    HallsofIvy

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    I would have done this slightly differently. The maximum and minimum values of sin x are 1 and -1 so the maximum and minimum values of this function are a+ c and c- a respectively. That is, the greatest height of the tip of the hand is a+ c and the least is c- a. The average of those is ((a+c)+(c-a))/2= c and that is the height of the center of the clock: 180 cm. a then, is the length of the second hand, 20 cm. At noon t= 0 so we have h(0)= 20 sin(nb)+ 180. But at noon, the second hand is straight up: 20 sin(nb)+ 180= 180+ 20 so sin(nb)= 1. That is, nb= [itex]pi/2[/itex]. Yes, the period is 60 seconds so [itex]n(60+ b)-nb= 2\pi[/itex] or [itex]60n= 2\pi[/itex]. [itex]n= \pi/30[/itex] and then [itex]b= (\pi/2)(30/\pi)= 2[/itex].
     
    Last edited: Aug 9, 2008
  6. Aug 9, 2008 #5

    Defennder

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    It looks like itex isn't working as it should.
     
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