Question

The displacement, (m), of a body in damped oscillation based on the following functions = 2− . (3) (8)

Answer 1

The equation for the first derivative dy/dt of the displacement function
`\(y = 2e^(-t) \sin(3t)\)` is
`\(y' = -2e^(-t)\sin(3t) + 6e^(-t)\cos(3t)\)`, obtained using the product rule.

Let
`\(y(t) = 2e^(-t) \sin(3t)\)` be the displacement function. To find dy/dt, the first derivative, we'll use the product rule.

The product rule states that if
`\(y = f(t)g(t)\)`, then
`\(y' = f'g + fg'\)`.

Let
`\(f(t) = 2e^(-t)\) and \(g(t) = \sin(3t)\)`. Then,

`\[y'(t) = f'(t)g(t) + f(t)g'(t)\]`

Now, find the derivatives:

`\[f'(t) = -2e^(-t)\]\[g'(t) = 3\cos(3t)\]`

Substitute these back into the product rule:

`\[y'(t) = (-2e^(-t))\sin(3t) + (2e^(-t))(3\cos(3t))\]`

So, the equation for y'(t) is:

`\[y'(t) = -2e^(-t)\sin(3t) + 6e^(-t)\cos(3t)\]`

The complete question is:

The displacement, y(m), of a body in damped oscillation is y = 2e^-t sin3t. use the product rule to find an equation for the object if y = dy/dt.