91Ó°ÊÓ

Damped harmonic motion is a type of vibrational motion where the amplitude decreases over time. This reduction in amplitude is often due to non-conservative forces such as friction or air resistance. In the context of this exercise, you have an object attached to a spring, which is displacing from its equilibrium position. The equation given for displacement, \(y = e^{-0.5t}(0.4 \cos 6t - 0.2 \sin 6t)\), describes how far the object is from equilibrium at any time \(t\). - The exponential term \(e^{-0.5t}\) represents the damping force. It causes the amplitude of the motion to decrease over time.- The trigonometric functions \(\cos\) and \(\sin\) model the periodic motion. They describe the oscillation of the object due to the restoring force of the spring.This type of motion is commonly observed in real-world scenarios where an oscillating system loses energy over time, like a swinging pendulum coming to rest.

The product rule is a fundamental calculus technique for differentiating functions that are the product of two other functions. In this exercise, the displacement function \(y\) is the product of two functions:- \(u(t) = e^{-0.5t}\)- \(v(t) = 0.4 \cos 6t - 0.2 \sin 6t\)To find the derivative of the displacement function, which gives the velocity of the object, we use the product rule. This rule states:\[(uv)' = u'v + uv'\]Where:- \(u'\) is the derivative of \(u\) with respect to \(t\)- \(v'\) is the derivative of \(v\) with respect to \(t\)This principle allows us to break down complex differentiation tasks into simpler parts. By differentiating each component separately and applying the rule, we find the derivative of the entire product-based function.

The derivative is a cornerstone concept in calculus that represents the rate of change of a function with respect to a variable. In this scenario, we need to differentiate the displacement function \(y\) with respect to time \(t\) to find the object's velocity. Differentiating means finding how quickly the displacement of the object changes over time. - The derivative of \(e^{-0.5t}\) with respect to \(t\) is \(-0.5e^{-0.5t}\). This involves applying the chain rule since it is an exponential function.- For the trigonometric part, using the chain rule: the derivative of \(0.4 \cos 6t\) is \(-2.4 \sin 6t\), and for \(-0.2 \sin 6t\), it is \(-1.2 \cos 6t\).By gathering these results, we can apply them using the product rule to find the velocity, which reveals how fast and in what direction the object is moving at any instant.

Velocity is a vector quantity that describes the rate of change of position. In this exercise, it is obtained by computing the derivative of the displacement function. The final simplified expression for the velocity \( y'(t) \) is given by:\[y'(t) = e^{-0.5t}(-1.4 \cos 6t - 2.5 \sin 6t)\]To find the velocity at a specific time, like at \(t = 0.26s\), substitute \(t = 0.26\) into this expression:\[y'(0.26) = e^{-0.5 \cdot 0.26}(-1.4 \cos(6 \cdot 0.26) - 2.5 \sin(6 \cdot 0.26))\]By using known values, approximate the result:- \(e^{-0.13} \approx 0.875\)- \(\cos(1.56) \approx 0.0\)- \(\sin(1.56) \approx 1.0\)Finally, calculate:\[y'(0.26) \approx -2.1875\]