Question

The motion of a spring that is subject to a frictional force or a damping force (such as a shock absorber in a car) is often modelled by the product of an exponential function and a sine or cosine function. Suppose the equation of motion of a point on such a spring is

\(s\left( t \right) = 2{e^{ - 1.5t}}\sin 2\pi t\)

where \(s\) is measured in centimetres and \(t\) in seconds. Find the velocity after \(t\) seconds and graph both the position and velocity functions for \(0 \le t \le 2\).

Short answer

The velocity after \(t\) seconds is \(2{e^{ - 1.5t}}\left( {2\pi \cos 2\pi t - 1.5\sin 2\pi t} \right)\).

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The product rule of differentiation

If a function\(h\left( x \right) = f\left( x \right) \cdot g\left( x \right)\)and\(f\)and\(g\)are both differentiable, then the derivative of\(h\left( x \right)\)is as follows:

\(\frac{{dh}}{{dx}} = \frac{d}{{dx}}\left( {f\left( x \right) \cdot g\left( x \right)} \right) = f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

The derivative of displacement

The velocity function is equivalent to the derivative of the displacement function. Differentiate the displacement function with respect to \(t\) as follows:

\(\begin{aligned}s'\left( t \right) &= 2\frac{d}{{dt}}\left( {{e^{ - 1.5t}}\sin 2\pi t} \right)\\v\left( t \right) &= 2\left( {{e^{ - 1.5t}}\frac{d}{{dt}}\left( {\sin 2\pi t} \right) + \sin 2\pi t\frac{d}{{dt}}\left( {{e^{ - 1.5t}}} \right)} \right)\\ &= 2\left( {{e^{ - 1.5t}}\left( {\cos 2\pi t} \right)\frac{d}{{dt}}\left( {2\pi t} \right) + \sin 2\pi t\left( {{e^{ - 1.5t}}} \right)\frac{d}{{dt}}\left( { - 1.5t} \right)} \right)\\ &= 2\left( {{e^{ - 1.5t}}\left( {\cos 2\pi t} \right)\left( {2\pi } \right) + \sin 2\pi t\left( {{e^{ - 1.5t}}} \right)\left( { - 1.5} \right)} \right)\\ &= 2{e^{ - 1.5t}}\left( {2\pi \cos 2\pi t - 1.5\sin 2\pi t} \right)\end{aligned}\)

Check the answer visually

The procedure to draw the graph of the above equation by using the graphing calculator is as follows:

To check the answer visually draw the graph of the function\(s\left( t \right) = 2{e^{ - 1.5t}}\sin 2\pi t\), \(v\left( t \right) = 2{e^{ - 1.5t}}\left( {2\pi \cos 2\pi t - 1.5\sin 2\pi t} \right)\)by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(2{e^{ - 1.5t}}\sin 2\pi t\)in the\({Y_1}\)tab.
2. Select the “STAT PLOT” and enter the equation\(2{e^{ - 1.5t}}\left( {2\pi \cos 2\pi t - 1.5\sin 2\pi t} \right)\)in the\({Y_2}\)tab.
3. Enter the “GRAPH” button in the graphing calculator.

Visualization of graphs of the functions\(s\left( t \right) = 2{e^{ - 1.5t}}\sin 2\pi t\), \(v\left( t \right) = 2{e^{ - 1.5t}}\left( {2\pi \cos 2\pi t - 1.5\sin 2\pi t} \right)\) are shown below: