A Function for Damped Simple Harmonic MotionDate: 10/17/98 at 23:36:08 From: William Liao Subject: Periodic and exponential function Dr. Maths, I have problem understanding the following problem in my textbook:: The end of a plastic ruler is vibrating in damped harmonic motion. When it starts vibrating, the amplitude is 0.9cm, but 2 seconds later has decreased to 0.4cm. In this time it has moved backwards and forwards 100 times. Find an expression for the position of the end of the ruler, taking the centre of vibration as 0. In the solution given for this sxample, I couldn't understand the part where it said: "In 2 seconds, 100 vibrations are completed, so the frequency is 50/second and the period is thus 0.02s, giving b = 50 x 2pi." (y = Ae(at) cos bt) Also I don't quite see how it said: "After 2 seconds the amplitude is only 0.4 cm, so 0.4 = 0.9e(2a)." Please try to explain it to me. Thank you. Sincerely, Will Date: 10/18/98 at 17:54:55 From: Doctor Anthony Subject: Re: Periodic and exponential function The equation for damped SHM (simple harmonic motion) is: x = Ae^(-kt)cos(wt) where A, k, w are constants to be found. When t = 0, x = 0.9 so 0.9 = A, and we have: x = 0.9e^(-kt)cos(wt) In 2 seconds it completes 100 oscillations and so does 50 cycles per second: Period = 2pi/w = 1/50 and w = 100pi x = 0.9e^(-kt)cos(100.pi.t) When t = 2, x = 0.4, so: 0.4 = 0.9e^(-2k)cos(200pi) 4/9 = e^(-2k) 2/3 = e^(-k) -k = ln(2/3) = -0.405 k = 0.405 Thus we get: x = 0.9e^(-0.405t)cos(100pi t) This is the equation of the damped simple harmonic motion. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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