The motion of a particle performing damped oscillations is given by the formula

y = e-t sin2t

Where y - displacement from its mean position and t - time in seconds


(a)determine the time at which the velocity of the particle is 0

(b)Determine if the displacement of the particle reaches a maximum or a minimum at the time when the velocity is 0

(c)Hence, find the displacement

what's the problem? As usual, take the drivative. y' is the velocity.

y' = -e^-t sin2t + 2e^-t cos2t
= e^-t (sin2t + 2cos2t)

so, velocity=0 when sin2t + 2cos2t = 0
sin2t = -2cos2t
tan2t = -2
so, find arctan(-2) and add multiples of pi/2 to t

To solve these questions, we need to find the velocity of the particle by first differentiating the given displacement equation with respect to time. Then we can answer the questions based on the calculated velocity.

Let's start with the given displacement equation:
y = e^-t sin(2t)

(a) To find the time at which the velocity of the particle is 0, we need to find when the derivative of y with respect to t is equal to zero.

First, let's differentiate y with respect to t:
dy/dt = d(e^-t sin(2t))/dt

To differentiate this equation, we can use the product rule, which states that the derivative of a product is the first term times the derivative of the second term plus the second term times the derivative of the first term.

Using this rule, we can differentiate the equation as follows:

dy/dt = -e^-t sin(2t) + e^-t * d(sin(2t))/dt

Next, we need to find d(sin(2t))/dt. The derivative of sin(2t) with respect to t is given by:

d(sin(2t))/dt = 2cos(2t)

Substituting this into our previous equation, we have:

dy/dt = -e^-t sin(2t) + 2e^-t cos(2t)

Now, set the derivative dy/dt equal to zero and solve for t:

0 = -e^-t sin(2t) + 2e^-t cos(2t)

At this point, we can divide the equation by e^-t to simplify it:

0 = -sin(2t) + 2cos(2t)

Rearranging the equation:

sin(2t) = 2cos(2t)

Now, we can solve for t. Divide both sides of the equation by cos(2t) to get:

tan(2t) = 2

To solve this equation, we can use inverse trigonometric functions. Taking the arctan of both sides:

2t = arctan(2)

Now, divide both sides by 2 to solve for t:

t = arctan(2) / 2

So, the time at which the velocity of the particle is zero is t = arctan(2) / 2.

(b) To determine if the displacement of the particle reaches a maximum or a minimum at the time when the velocity is zero, we need to find the second derivative of the displacement equation.

The second derivative gives us information about the concavity of the function. If the second derivative is positive, the displacement has a minimum at that time. If the second derivative is negative, the displacement has a maximum.

To find the second derivative, we differentiate dy/dt:

d^2y/dt^2 = d^2(e^-t sin(2t))/dt^2

Using the product rule again, we differentiate the equation as follows:

d^2y/dt^2 = d(-e^-t sin(2t))/dt + d(e^-t * d(sin(2t))/dt)/dt

Differentiating the terms:

= e^-t sin(2t) - 2e^-t cos(2t) - e^-t * d(cos(2t))/dt

Now, differentiate d(cos(2t))/dt:

= e^-t sin(2t) - 2e^-t cos(2t) + e^-t * d(sin(2t))/dt

= e^-t sin(2t) - 2e^-t cos(2t) + e^-t * 2cos(2t)

= e^-t sin(2t) - 2e^-t cos(2t) + 2e^-t cos(2t)

Simplifying the equation:

= e^-t sin(2t)

Now, substitute t = arctan(2) / 2 into the equation:

= e^(-arctan(2)/2) sin(2 * arctan(2) / 2)

To determine if the displacement reaches a maximum or a minimum at this time, we need to evaluate the sign of this expression.

(c) Finally, to find the displacement at this time, substitute t = arctan(2) / 2 into the original displacement equation:

y = e^(-arctan(2)/2) sin(2 * arctan(2) / 2)