document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1171811>Question 1171811</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An object moves in simple harmonic motion described by the equation d=5 cos pit​, where t is measured in seconds and d in inches. Find the following.<BR>\n" );
document.write( "a. the maximum displacement <BR>\n" );
document.write( "b. the frequency <BR>\n" );
document.write( "c. the time required for one cycle\r<BR>\n" );
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document.write( "I need to find a,b,c </SPAN>\n" );
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document.write( "Part (a)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d+=+5cos%28pi%2At%29 ALIGN=MIDDLE  ALT=\"d+=+5cos%28pi%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the same as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+5cos%28pi%2Ax%29 ALIGN=MIDDLE  ALT=\"y+=+5cos%28pi%2Ax%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > which can be written into <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+5cos%28pi%28x-0%29%29%2B0 ALIGN=MIDDLE  ALT=\"y+=+5cos%28pi%28x-0%29%29%2B0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Comparing that last equation to the form <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+Acos%28B%28x-C%29%29%2BD ALIGN=MIDDLE  ALT=\"y+=+Acos%28B%28x-C%29%29%2BD\"   DISABLED_style= \"max-width:90%; height:auto;\" >, we have:<BR>\n" );
document.write( "A = 5<BR>\n" );
document.write( "B = pi<BR>\n" );
document.write( "C = 0<BR>\n" );
document.write( "D = 0<BR>\n" );
document.write( "Note: we won't be using the values of C and D for this problem (they handle horizontal and vertical shifting).\r<BR>\n" );
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document.write( "The value of A determines the amplitude. We have |A| = |5| = 5 as our amplitude which is the maximum displacement. The object moves up and down, going at most 5 inches above or below its initial position. I recommend graphing <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+5cos%28pi%2Ax%29 ALIGN=MIDDLE  ALT=\"y+=+5cos%28pi%2Ax%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > to see this.\r<BR>\n" );
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document.write( "Answer: 5 inches. \r<BR>\n" );
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document.write( "Part (b)\r<BR>\n" );
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document.write( "The value of B will help us find the frequency. First we compute the period T\r<BR>\n" );
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document.write( "T = 2pi/B<BR>\n" );
document.write( "T = 2pi/pi<BR>\n" );
document.write( "T = 2\r<BR>\n" );
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document.write( "The period is 2 seconds. This means every 2 seconds, the cycle repeats itself. In other words, the length of each cycle is 2 seconds.\r<BR>\n" );
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document.write( "The frequency f is 1 over the period, aka the reciprocal of the period.\r<BR>\n" );
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document.write( "frequency = 1/(period)<BR>\n" );
document.write( "f = 1/T<BR>\n" );
document.write( "f = 1/2<BR>\n" );
document.write( "f = 0.5\r<BR>\n" );
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document.write( "The frequency is 1/2 = 0.5 cycles per second. Every second, the object completes half a cycle. Note how the units for T are in \"seconds per cycle\". When we apply the reciprocal, the units swap getting \"cycles per second\". \r<BR>\n" );
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document.write( "Answer: 1/2 = 0.5 cycles per second.\r<BR>\n" );
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document.write( "Part (c)\r<BR>\n" );
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document.write( "As stated earlier in part (b), the period is 2 seconds, which is the length of one full cycle. We don't have much else to do here.\r<BR>\n" );
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document.write( "Answer: 2 seconds<BR>\n" );
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