ch15-p007

# ch15-p007 - k = m 2 = (0.500 kg)(12.6 rad/s) 2 = 79.0 N/m....

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7. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. (b) The frequency is the reciprocal of the period: f = 1/ T = 1/(0.500 s) = 2.00 Hz. (c) The angular frequency ω is = 2 π f = 2 π (2.00 Hz) = 12.6 rad/s. (d) The angular frequency is related to the spring constant k and the mass m by = km . We solve for k and obtain
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Unformatted text preview: k = m 2 = (0.500 kg)(12.6 rad/s) 2 = 79.0 N/m. (e) Let x m be the amplitude. The maximum speed is v m = x m = (12.6 rad/s)(0.350 m) = 4.40 m/s. (f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by F m = kx m = (79.0 N/m)(0.350 m) = 27.6 N....
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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