# displacement (s) of a spring is given as s=ae^(-kt)sin2pi*f*t where t is time.

determine velocity and acceleration affter 2seconds if: a=3, k=0.75 and f=20.

i think im supposed to integrate but cant think how to do this one.

## Well, I'm no spring expert, but let's give it a shot! To find velocity and acceleration, we need to differentiate the displacement equation with respect to time, which means finding the derivative.

Starting with s = ae^(-kt)sin(2πft), let's tackle it one step at a time.

The derivative of e^(-kt) is -ke^(-kt) (using the chain rule).

The derivative of sin(2πft) with respect to t is simply 2πf * cos(2πft) (by the chain rule).

Now, we can differentiate the entire equation by putting these parts together:

ds/dt = -kae^(-kt)sin(2πft) - 2πf * ae^(-kt)cos(2πft).

This expression gives us the velocity, which represents how fast the displacement is changing over time.

Next, to find acceleration, we differentiate the velocity equation once more with respect to time:

d^2s/dt^2 = (-k)^2ae^(-kt)sin(2πft) - 2πfkae^(-kt)cos(2πft) - (2πf)^2ae^(-kt)sin(2πft).

Now we have the equation for acceleration, which represents the rate of change of velocity.

To find the velocity and acceleration at t = 2 seconds, simply plug in the values of a = 3, k = 0.75, f = 20, and t = 2 into the equations above. You might need a calculator for this step, or perhaps you're more mathematically inclined than I am!

Remember, these equations are just an approximation, and there might be additional factors or physics to consider. But hey, sometimes physics equations can be a bit like a clown juggling balls – a tricky balancing act!

## To find the velocity and acceleration, we need to differentiate the displacement equation with respect to time.

Given: s = ae^(-kt)sin(2πft)

1. Velocity (v):

To find the velocity, we need to differentiate the displacement equation with respect to time.

v = ds/dt

First, let's find ds/dt of each term separately.

a. Differentiating ae^(-kt):

To differentiate ae^(-kt), we will use the chain rule.

d/dt (ae^(-kt)) = -ak * e^(-kt)

b. Differentiating sin(2πft):

To differentiate sin(2πft), we will use the chain rule.

d/dt (sin(2πft)) = 2πf * cos(2πft)

Combining these results, we have:

v = -ak * e^(-kt) * sin(2πft) + a * e^(-kt) * 2πf * cos(2πft)

2. Acceleration (a):

To find the acceleration, we need to differentiate the velocity equation with respect to time.

a = dv/dt

To differentiate the velocity equation, we will use the product rule.

Differentiating the first term, -ak * e^(-kt) * sin(2πft):

d/dt (-ak * e^(-kt) * sin(2πft)) = -ak * (-k) * e^(-kt) * sin(2πft) + -ak * e^(-kt) * (2πf) * cos(2πft)

Differentiating the second term, a * e^(-kt) * 2πf * cos(2πft):

d/dt (a * e^(-kt) * 2πf * cos(2πft)) = a * (-k) * e^(-kt) * 2πf * cos(2πft) + a * e^(-kt) * 2πf * (-2πf) * sin(2πft)

Combining these results, we have:

a = ak^2 * e^(-kt) * sin(2πft) - 2akπf * e^(-kt) * cos(2πft) - ak * e^(-kt) * (2πf)^2 * sin(2πft)

Now, we have the equations for velocity (v) and acceleration (a) based on the given displacement equation.

To find the velocity and acceleration after 2 seconds, substitute the values of a = 3, k = 0.75, f = 20, and t = 2 seconds into the equations for v and a, respectively.

## To find the velocity and acceleration of the spring after 2 seconds, you need to differentiate the displacement equation with respect to time.

Given: s = ae^(-kt)sin(2πft)

Where:

a = 3 (amplitude)

k = 0.75 (damping constant)

f = 20 (frequency)

The displacement equation is: s = ae^(-kt)sin(2πft)

1. To find the velocity, differentiate the displacement equation with respect to time (t):

v = ds/dt

Using the chain rule, we need to differentiate each term in the equation.

a. Differentiating ae^(-kt):

We can use the chain rule here:

dy/dt = (dy/du)*(du/dt)

Let u = -kt, so du/dt = -k

dy/du = a * e^u

dy/dt = a * e^(-kt) * (-k)

b. Differentiating sin(2πft):

The derivative of sin(2πft) with respect to t is:

d(sin(2πft))/dt = 2πf * cos(2πft)

Putting it all together:

v = a * e^(-kt) * (-k) * sin(2πft) + 2πf * a * e^(-kt) * cos(2πft)

2. To find the acceleration, differentiate the velocity equation with respect to time (t):

a = dv/dt

Using the chain rule again, differentiate each term in the velocity equation:

a. Differentiating (a * e^(-kt) * (-k) * sin(2πft)):

We can use the chain rule here:

dy/dt = (dy/du)*(du/dt)

Let u = -kt, so du/dt = -k

dy/du = a * e^u

dy/dt = a * e^(-kt) * (-k)

b. Differentiating (2πf * a * e^(-kt) * cos(2πft)):

The derivative of 2πf * a * e^(-kt) * cos(2πft) with respect to t is:

d(2πf * a * e^(-kt) * cos(2πft))/dt = -2πf * a * e^(-kt) * sin(2πft)

Putting it all together:

a = a * e^(-kt) * (-k)^2 * sin(2πft) + 2πf * a * e^(-kt) * cos(2πft) - 2πf * a * e^(-kt) * sin(2πft)

Now that you have the equations for velocity (v) and acceleration (a), you can substitute the given values of a = 3, k = 0.75, and f = 20, and the time t = 2 into the equations to get the values of velocity and acceleration after 2 seconds.

## s = ae^-kt sin 2πft

s' = -ak e^-kt sin 2πft + ae^-kt (2πf) cos 2πft

= ae^-kt (-ksin 2πft + 2πf cos 2πft)

s" = -ake^-kt (-ksin 2πft + 2πf cos 2πft) + ae^-kt (-2πkf sin 2πft - 4π^2f^2 sin 2πft)

I'll let you check my algebra and plug in the values.