document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=32326>Question 32326</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The period (the time required for one complete swing) of a simple pendulum varies directly as the square root of its length.  If a pendulum 12 feet long has a period of 4 seconds, find the period of a pendulum 3 feet long. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #18945 by </B><A HREF=/tutors/aboutme.mpl?userid=mukhopadhyay><B>mukhopadhyay(490)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mukhopadhyay><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=18945\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Period(T) = k(Length)^1/2<BR>\n" );
document.write( "4 = k(12)^1/2<BR>\n" );
document.write( "=>4=2k(sqrt3)<BR>\n" );
document.write( "=>k=2/sqrt(3)<BR>\n" );
document.write( "So, T = 2L/sqrt(3)<BR>\n" );
document.write( "If L=3<BR>\n" );
document.write( "T=6/sqrt(3) = 2sqrt(3) = 3.5 seconds(approx)<BR>\n" );
document.write( "Answer: approx 3.5 seconds </SPAN>\n" );
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