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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 are 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.15.18 by Eq.15.17, we eliminate one of those unknowns and reduce the other to a single trignometric function: `(v(0))/(x(0)) = (-omega x_(m) sin phi)/(x_(m) cos phi) = - omega TAN phi` Solving for `tan phi`, we find `tan phi = -(v(0))/(omega x(0))= - (-0.920m//s)/((23.5"rad/s")(-0.0850m))` `= -0.461` This equation has TWO solutions: `phi = -25^(@) and phi = -180^(@) + (-25^(@)) = 155^(@)`. Normally only the first solution here is displayed by a calculator, but it may not be the physically possible solution. To choose the proper solution, we TEST them both by using them to compute values for the AMPLITUDE `x_(m)`. From Eq. 15.17, we find 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 positive constant, the correct phase constant and amplitude here are `phi = -155^(@) and x_(m) = 0.094m = 9.4cm` |
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