Physics, 01.12.2019 03:31 thereal6ix9ine69
X=16.2 cos(4.68t+0.420). what is the velocity of the oscillator at t=7.97 s?
Answers: 1
would be equal to approximately 
.
here stands for the displacement of the oscillator from its equilibrium position at time 
. ![v(t) = \dfrac{d}{dt} [x(t)] = \dfrac{d}{dt} [16.2\, \cos(4.68 \, t + 0.420)]= 16.2 \, \dfrac{d}{dt} [16.2\, \cos(4.68 \, t + 0.420)]](/tpl/images/0397/8988/86067.png)
. ![\dfrac{d}{dt} \,[\cos(f(t))] = - \sin(t) \cdot \dfrac{d}{dt}[f(t)]](/tpl/images/0397/8988/57884.png)
by the chain rule. In this case the inner function is 
. Its first derivative is equal to 
. Hence![\dfrac{d}{dt} [\cos(4.68 \, t + 0.420)] = - \sin(4.68\, t + 0.420) \cdot 4.68 = -4.68\, \sin(4.68\, t + 0.420)](/tpl/images/0397/8988/61ba2.png)
. 
.
, that would be equal to 
.
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X=16.2 cos(4.68t+0.420). what is the velocity of the oscillator at t=7.97 s?...
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