411_PartUniversity Physics Solution

# 411_PartUniversity Physics Solution - Periodic Motion...

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13-4 Chapter 13 EVALUATE: We can also calculate the maximum force from the maximum acceleration: 2 max max (0.253 kg)(12.0 m/s ) 3.04 N, Fm a == = which agrees with our previous results. 13.18. IDENTIFY: The general expression for ( ) x vt is ( ) sin( ) x A t ω φ = −+ . We can determine and A by comparing the equation in the problem to the general form. SET UP: 4.71 rad/s = . 3.60 cm/s 0.0360 m/s A . EXECUTE: (a) 22 r a d 1.33 s 4.71 rad/s T ππ = (b) 3 0.0360 m/s 0.0360 m/s 7.64 10 m 7.64 mm 4.71 rad/s A === × = (c) 3 2 max (4.71 rad/s) (7.64 10 m) 0.169 m/s aA × = (d) k m = so (0.500 kg)(4.71 rad/s) 11.1 N/m km = .
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411_PartUniversity Physics Solution - Periodic Motion...

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