Question

The displacement of a particle on a vibrating string is given by the equation \(s\left( t \right) = 10 + \frac{1}{4}\sin \left( {10\pi t} \right)\) where \(s\) is measured in centimeters and \(t\) in seconds. Find the velocity of the particle after \(t\) seconds.

Short answer

The velocity of the particle after \(t\) seconds is \(v\left( t \right) = \frac{{5\pi }}{2}\cos \left( {10\pi t} \right)\)cm/s.

Step-by-step solution

The Chain Rule

The chain rule is defined as:

\(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\)

By using Leibniz notation,

\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

Determine the velocity of the particle after t seconds

Use the chain rule to differentiate the equation \(s\left( t \right)\) to obtain the velocity of the particle as shown below:

\(\begin{aligned}v\left( t \right) &= s'\left( t \right)\\ &= \frac{d}{{dt}}\left( {10 + \frac{1}{4}\sin \left( {10\pi t} \right)} \right)\\ &= 0 + \frac{1}{4}\cos \left( {10\pi t} \right) \cdot \frac{d}{{dt}}\left( {10\pi t} \right)\\ &= \frac{1}{4}\cos \left( {10\pi t} \right) \cdot \left( {10\pi } \right)\\ &= \frac{{5\pi }}{2}\cos \left( {10\pi t} \right)\end{aligned}\)

Thus, the velocity of the particle after \(t\) seconds is \(v\left( t \right) = \frac{{5\pi }}{2}\cos \left( {10\pi t} \right)\) cm/s.