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andriy [413]
4 years ago
5

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

Physics
2 answers:
pochemuha4 years ago
7 0

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

Leni [432]4 years ago
7 0

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.

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