ch15-p022 - 22. Eq. 15-12 gives the angular velocity: = k...

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22. Eq. 15-12 gives the angular velocity: 100 N/m 7.07rad/s. 2.00 kg k m ω == = Energy methods (discussed in §15-4) provide one method of solution. Here, we use trigonometric techniques based on Eq. 15-3 and Eq. 15-6. (a) Dividing Eq. 15-6 by Eq. 15-3, we obtain =+ v x t ωω φ tan b g so that the phase ( t + ) is found from () ( ) 11 3.415 m/s tan tan . 7.07 rad/s 0.129 m v t x ωφ −− §· += = ¨¸ ©¹ With the calculator in radians mode, this gives the phase equal to –1.31 rad. Plugging this back into Eq. 15-3 leads to0.129m cos( 1.31)
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