document.write( "Question 78467: 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 1 second period?
\n" ); document.write( "a. 1 foot
\n" ); document.write( "b. 1.5 foot
\n" ); document.write( "c. 2 feet
\n" ); document.write( "d. 3 feet
\n" ); document.write( "

Algebra.Com's Answer #56275 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
We have the equation\r
\n" ); document.write( "\n" ); document.write( "\"T=k%2Asqrt%28L%29\" \"The period of a simple pendulum is directly proportional to the square root of its length\"\r
\n" ); document.write( "\n" ); document.write( "Substitute L=6 and T=2 and solve for k\r
\n" ); document.write( "\n" ); document.write( "\"2=k%2Asqrt%286%29\"
\n" ); document.write( "\"2%2Fsqrt%286%29=k%2Across%28sqrt%286%29%2Fsqrt%286%29%29\" Divide both sides by \"sqrt%286%29\"\r
\n" ); document.write( "\n" ); document.write( "So our constant is \r
\n" ); document.write( "\n" ); document.write( "\"k=2%2Fsqrt%286%29\"\r
\n" ); document.write( "\n" ); document.write( "Now the equation becomes\r
\n" ); document.write( "\n" ); document.write( "\"T=2%2Fsqrt%286%29%2Asqrt%28L%29\"\r
\n" ); document.write( "\n" ); document.write( "Now plug in T=1 to solve for L\r
\n" ); document.write( "\n" ); document.write( "\"1=2%2Fsqrt%286%29%2Asqrt%28L%29\"
\n" ); document.write( "\"1%2F%282%2Fsqrt%286%29%29=sqrt%28L%29\" Divide both sides by \"2%2Fsqrt%286%29\"
\n" ); document.write( "\"sqrt%286%29%2F2=sqrt%28L%29\" Divide
\n" ); document.write( "\"%28sqrt%286%29%2F2%29%5E2=%28sqrt%28L%29%29%5E2\" Square both sides
\n" ); document.write( "\"6%2F4=L\"
\n" ); document.write( "\"L=3%2F2\" reduce\r
\n" ); document.write( "\n" ); document.write( "So our length must be 1.5 feet to have a period of 1 second.\r
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