Harmonic Stage Platform Vibration (Entertainment Engineering) Solution

Problem Overview

We evaluate the instantaneous velocity $v(t) = y'(t)$ of a dynamic stage lift given the coupled position function:

y(t) = (3t + 2)\cos(t)

Step-by-Step Mathematical Solution

Step 1: Structural Rule Identification and Component Dictionary

The outermost operation is a product between the linear polynomial $(3t + 2)$ and the trigonometric function $\cos(t)$ . The Product Rule ( $\frac{d}{dt}[uv] = u'v + uv'$ ) governs this derivative:

$u(t) = 3t + 2 \implies u'(t) = 3$
$v(t) = \cos(t) \implies v'(t) = -\sin(t)$

Step 2: Differentiate via Product Rule

v(t) = y'(t) = u'(t)v(t) + u(t)v'(t)

v(t) = 3\cos(t) + (3t + 2)(-\sin(t))

v(t) = 3\cos(t) - (3t + 2)\sin(t)

Step 3: Evaluate Velocity at $t = \pi\text{ seconds}$

Substitute $t = \pi$ :

$\cos(\pi) = -1$
$\sin(\pi) = 0$

v(\pi) = 3(-1) - (3\pi + 2)(0) = -3\text{ m/s}

Engineering Conclusion

At $t = \pi\text{ seconds}$ , the stage platform is moving downward at a velocity of $3\text{ m/s}$ . The negative sign indicates downward travel as the oscillatory drive passes through the negative half of its cycle. Stage automation controllers monitor this velocity to ensure hydraulic actuator valves do not exceed maximum flow rate limits.

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