```Question 54072
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>
```