Answer:
Assuming that the angle is in radians, the velocity of the oscillator at [tex]t = \rm 7.97 \; s[/tex] would be equal to approximately [tex]\rm -1.55[/tex].
Explanation:
Assume that the [tex]x[/tex] here stands for the displacement of the oscillator from its equilibrium position at time [tex]t[/tex].
Velocity is the first derivative of displacement with respect to time. So is the case in this question. Differentiate the expression for [tex]x[/tex] with respect to [tex]t[/tex] to find the velocity at time [tex]t[/tex]:
[tex]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)][/tex].
In calculus, [tex]\dfrac{d}{dt} \,[\cos(f(t))] = - \sin(t) \cdot \dfrac{d}{dt}[f(t)][/tex] by the chain rule. In this case the inner function is [tex]f(t) = 4.68 \, t + 0.420[/tex]. Its first derivative is equal to [tex]f^{\prime}(t) = 4.68[/tex]. Hence
[tex]\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) [/tex].
Therefore
[tex]v(t) = -(4.68 \times 16.2) \, \sin(4.68 \, t + 0.420)[/tex].
At time [tex]t = \rm 7.97\; s[/tex], that would be equal to
[tex]-16.2 \times 4.68 \cdot \sin(4.68 \times 7.97 + 0.420) \approx \rm 1.55[/tex].
Hence the (linear) velocity of the oscillator at [tex]t = \rm 7.97 \; s[/tex] would be equal to [tex]\rm -1.55[/tex].
