Question 54072
<pre><font face = "lucida console" size = 5 color = "indigo"><b>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 = k<font face = "symbol">Ö</font>Length
           _   
     P = k<font face = "symbol">Ö</font>L

Substitute L = 6 and P = 2 and solve for k
           _  
     2 = k<font face = "symbol">Ö</font>6
                      _
Divide both sides by <font face = "symbol">Ö</font>6

     2
   --<u>-</u>- = k
    <font face = "symbol">Ö</font>6

Rationalize the denominator
         _
     2  <font face = "symbol">Ö</font>6
    --<u>-</u>--<u>-</u>- = k
     <font face = "symbol">Ö</font>6 <font face = "symbol">Ö</font>6
        _
      2<font face = "symbol">Ö</font>6
     ----- = k
       6

Cancel 2 into 6
      _
     <font face = "symbol">Ö</font>6
    ---- = k
      3

Now substitute

        _
       <font face = "symbol">Ö</font>6
  k = ---- 
       3

into the original equation

           _   
     P = k<font face = "symbol">Ö</font>L

        _ _
       <font face = "symbol">Ö</font>6<font face = "symbol">Ö</font>L
  P = ---- 
        3
        __
       <font face = "symbol">Ö</font>6L
  P = ------ 
        3

Now substitute P = 1

        __
       <font face = "symbol">Ö</font>6L
  1 = ------ 
        3

Multiply both sides by 3 to clear of fractions
       __
  3 = <font face = "symbol">Ö</font>6L 
   
Square both sides of the equation

  9 = 6L

Divide both sides by 6

9/6 = L

3/2 = L

  L = 3/2   
  
Edwin</pre>