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At t=0, the displacement x(0) of the block in a linear oscillator like that of Fig. 15.5 is -8.50cm. (Read x(0) as ''x at time zero''). The block's velocity v(0) then is -0.920m//s, and its acceleration a(0) is +47.0 m//s^(2). What is the phase constant phi and amplitude x_(m) ? |
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Answer» SOLUTION :Calculations: We know `omega` and want `phi and x_(m)`. If we divide Eq. 15.26 by Eq. 15.25, we find `(v(0))/(x(0)) = (-omega x_(m) sin phi)/(x_(m) cos phi)` Solving for `tan phi`, we find `tan phi = - (v(0))/(omega x(0)) = - (-0.920m//s)/((23.5 "rad/s") (-0.0850m))` This EQUATION has two solutions: `phi = -25^(@) and phi = 180^(@) + (-25^(@)) = 155^(@)` (Normally, only the FIRST solution here is displayed by a calculator), To choose the proper solution, we test them both by using them to compute values for the amplitude `x_(m)`, From Eq. 15.25, we find that if `phi = -25^(@)`, then `x_(m) = (x(0))/(cos phi) = (-0.0850m)/(cos (-25^(@))) = - 0.094m` We find similarly that if `phi = 155^(@), " then " x_(m) = 0.094m`. Because the amplitude of SHM must be a positive constant, the correct phase constant and amplitude here are `phi = 155^(@) and x_(m) = 0.094m = 9.4cm` |
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