SOLUTION: The period of vibration P for a pendulum varies directly as the square root of the length L. If the period of vibration is 3 seconds when the length is 36 inches, what is the perio

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Question 127060: The period of vibration P for a pendulum varies directly as the square root of the length L. If the period of vibration is 3 seconds when the length is 36 inches, what is the period when L=5.0625 inches?
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Let P represent the "period of vibration." And L represent the length. Since the period of
vibration varies directly as the square root of the length, we can write the formula as:
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P+=+K%2Asqrt%28L%29
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K is called the "constant of proportionality." It is a constant that multiplies the square
root of L to make the answer equal to P. Since K is a constant, we can tell from the equation
that if L increases, then P must increase also to keep both sides of the equation equal.
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The first question is, "What is the value of K?"
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From the information in the problem we know that when P = 3, then L = 36. Substitute these
two values into the equation and you have:
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3+=+K%2Asqrt%2836%29
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But the square root of 36 is 6. Replace the square root of 36 by 6 and the equation becomes:
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3+=+K%2A6
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Solve for K by dividing both sides of this equation by 6 to get:
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3%2F6+=+K
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And simplifying the fraction we get:
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K+=+1%2F2
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Now that we have K we can substitute 1%2F2 for K in the general equation and we get:
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P+=+%281%2F2%29%2Asqrt%28L%29
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Now to solve for any period all we have to do is to substitute the Length. For this problem
we are told to find the "period of vibration" when L= 5.0625 inches. Substitute this value of
L into the equation and you have:
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P+=+%281%2F2%29%2Asqrt%285.0625%29
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Take the square root of 5.0625 and you get 2.25. Then substitute that value into the equation
in place of sqrt%285.0625%29 and the equation becomes:
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P+=+%281%2F2%29%2A2.25
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Multiply out the right side and you have:
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P+=+1.125
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So the answer to this problem is the period of vibration is 1.125 seconds when the length
of the arm of the pendulum is 5.0625 inches.
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Hope this helps you to understand the problem and how to work it.
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