Integrator Design for Acceleration of Falling Object | Op Amps Homework Solution

[graviton123]

Homework Statement

Hi! I've to try and solve "use two opamp integrators to solve the differental acceleration of an object falling under the force of gravity d^2y/dt^2 = 10"

Homework Equations

Vout = -1/Rc * integral (Vin dt)

The Attempt at a Solution

I'm really stuck as to how to approach this! I've gotten as far as the voltage being potential difference and the work can be given by force * distance and RC is the time constant for a capacitor but i don't know where to go next if anyone could offer some help itd be great!
thanks

 

[mark726]

Dear student,

It seems like you have a good understanding of the basic concepts involved in this problem. To solve the differential acceleration of an object falling under the force of gravity, we can use two opamp integrators. Let's break down the problem step by step.

First, let's define our variables. We have the acceleration (a), which is equal to 10 m/s^2 in this case. We also have the initial velocity (v0) and initial position (y0) of the object.

Next, we can use the equation for acceleration to find the velocity (v) and position (y) of the object at any given time (t):
a = dv/dt
v = at + v0
y = 1/2at^2 + v0t + y0

Now, let's focus on the opamp integrators. As you mentioned, the output voltage of an integrator is given by Vout = -1/Rc * integral (Vin dt). In our case, Vin is the input voltage, which is equal to the acceleration (a) in this problem. So, we can write:
Vout = -1/Rc * integral (a dt)

To solve this integral, we can use the equations for velocity and position that we derived earlier. Substituting these equations into the integral, we get:
Vout = -1/Rc * integral (at + v0 dt) = -1/Rc * (1/2at^2 + v0t + y0)

This is the output voltage of one of the opamp integrators. To get the output voltage of the second integrator, we can simply cascade the two integrators. This means connecting the output of the first integrator to the input of the second integrator. The output voltage of the second integrator will then be:
Vout2 = -1/Rc * integral (Vout dt)

Substituting the equation for Vout that we derived earlier, we get:
Vout2 = -1/Rc * integral (-1/Rc * (1/2at^2 + v0t + y0) dt)

Solving this integral, we get:
Vout2 = 1/Rc^2 * (1/6at^3 + 1/2v0t^2 + y0t) + C

Where C is a constant of integration. This is the final output