Question

X=16.2 cos(4.68t+0.420). What is the velocity of the oscillator at t=7.97 s?

Answer 1

Answer:

If using radians: 16.2

If using degrees: 12.8

Don't forget your units

Explanation:

I assume that x represents the velocity?

If so, then just substitute the value t and solve

Answer 2

Answer:

Assuming that the angle is in radians, the velocity of the oscillator at $t = \rm 7.97 \; s$ would be equal to approximately $\rm -1.55$.

Explanation:

Assume that the $x$ here stands for the displacement of the oscillator from its equilibrium position at time $t$.

Velocity is the first derivative of displacement with respect to time. So is the case in this question. Differentiate the expression for $x$ with respect to $t$ to find the velocity at time $t$:

$v(t) = \dfrac{d}{dt} [x(t)] = \dfrac{d}{dt} [16.2\, \cos(4.68 \, t + 0.420)]= 16.2 \, \dfrac{d}{dt} [16.2\, \cos(4.68 \, t + 0.420)]$.

In calculus, $\dfrac{d}{dt} \,[\cos(f(t))] = - \sin(t) \cdot \dfrac{d}{dt}[f(t)]$ by the chain rule. In this case the inner function is $f(t) = 4.68 \, t + 0.420$. Its first derivative is equal to $f^{\prime}(t) = 4.68$. Hence

$\dfrac{d}{dt} [\cos(4.68 \, t + 0.420)] = - \sin(4.68\, t + 0.420) \cdot 4.68 = -4.68\, \sin(4.68\, t + 0.420)$.

Therefore

$v(t) = -(4.68 \times 16.2) \, \sin(4.68 \, t + 0.420)$.

At time $t = \rm 7.97\; s$, that would be equal to

$-16.2 \times 4.68 \cdot \sin(4.68 \times 7.97 + 0.420) \approx \rm 1.55$.

Hence the (linear) velocity of the oscillator at $t = \rm 7.97 \; s$ would be equal to $\rm -1.55$.