SOLUTION: Please help me with this problem: The period of a simple pendulum is directly proportional to the square root of its length. If a pendulum has a length of 6 feet and a period of 2

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Question 54072This question is from textbook Algebra and Trigonometry
: Please help me with this problem:
The period of a simple pendulum is directly proportional to the square root of its length. If a pendulum has a length of 6 feet and a period of 2 seconds, to what length should it be shortened to achieve a 1 second period? This question is from textbook Algebra and Trigonometry

Found 2 solutions by stanbon, AnlytcPhil:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The period of a simple pendulum is directly proportional to the square root of its length.
This means Period = ksqrt(length)
You need to find "k".
------
If a pendulum has a length of 6 feet and a period of 2 seconds
This means 2=ksqrt6
Then k=2/sqrt6
--------
So now we know the formula for a perdulum is:
P=(2/sqrt6)sqrt(length)
P=2sqrt(length/6)
---------
To what length should it be shortened to achieve a 1 second period?
Using that formula you get:
1=2sqrt(length/6)
sqrt(length/6)=1/2
Square both sides to get:
(length/6)=1/4
length = 3/2 ft
-------
Cheers,
Stan H.

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me with this problem:
The period of a simple pendulum is directly 
proportional to the square root of its length. 
If a pendulum has a length of 6 feet and a 
period of 2 seconds, to what length should it 
be shortened to achieve a 1 second period?

           ______
Period = kÖLength
           _   
     P = kÖL

Substitute L = 6 and P = 2 and solve for k
           _  
     2 = kÖ6
                      _
Divide both sides by Ö6

     2
   ---- = k
    Ö6

Rationalize the denominator
         _
     2  Ö6
    ------- = k
     Ö6 Ö6
        _
      2Ö6
     ----- = k
       6

Cancel 2 into 6
      _
     Ö6
    ---- = k
      3

Now substitute

        _
       Ö6
  k = ---- 
       3

into the original equation

           _   
     P = kÖL

        _ _
       Ö6ÖL
  P = ---- 
        3
        __
       Ö6L
  P = ------ 
        3

Now substitute P = 1

        __
       Ö6L
  1 = ------ 
        3

Multiply both sides by 3 to clear of fractions
       __
  3 = Ö6L 
   
Square both sides of the equation

  9 = 6L

Divide both sides by 6

9/6 = L

3/2 = L

  L = 3/2   
  
Edwin