document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=180018>Question 180018</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The frequency of a vibrating string varies as its length. A string 3 feet long vibrates 175 cycles per second. Find the frequency of a 5 foot string. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #134939 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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/cgi-bin/show-face.mpl?user=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=134939\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The frequency of a vibrating string varies as its length. A string 3 feet long vibrates 175 cycles per second. Find the frequency of a 5 foot string.<BR>\n" );
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document.write( "Actually, it's an inverse relationship.  A longer string will vibrate at a lower frequency (all other factors being equal).<BR>\n" );
document.write( "So, 175 = k/3 (k is a constant)<BR>\n" );
document.write( "k = 525<BR>\n" );
document.write( "f2 = 525/5<BR>\n" );
document.write( "f2 = 105 Hz (Hz, Hertz, is cycles per second)<BR>\n" );
document.write( "f2 of the 5' string is 105 Hz<BR>\n" );
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