Given that x is the displacement and v is the velocity and F...

Given that x is the displacement and v is the velocity and F is the input force the equation for the M2 equation is: Select one: a. $M_2 \frac{d^2x_2}{dt^2} + B_2 \frac{dx_2}{dt} + K_2x_2 + B_{12} \frac{d(x_2+x_1)}{dt} + K_1(x_2 + x_1) = 0$ b. $M_2 \frac{d^2x_2}{dt^2} + B_2 \frac{dx_2}{dt} + B_{12} \frac{d(x_2-x_1)}{dt} + K_1(x_2 - x_1) = 0$ c. $M_2 \frac{d^2x_2}{dt^2} + B_2 \frac{dx_2}{dt} + B_{12} \frac{d(x_2+x_1)}{dt} + K_1 (x_2 + x_1) = 0$ d. $M_2 \frac{d^2x_2}{dt^2} + B_2 \frac{dx_2}{dt} + K_2x_2 + B_{12} \frac{d(x_2+x_1)}{dt} + K_1(x_2 - x_1) = 0$

Transcript

00:01 Hello students, in this question we have given the figure that is x and this is kx or bx dot.

00:13 Now this is due to damper.

00:16 Now first we have to determine the differential equation for the system to find using from newton's second law that is f is equal to ma so we can write here mx is equal to ft minus kx minus bx.

00:41 Now this will be written as mx plus bx plus kx will be equal to ft.

00:52 Now second second in this we have to determine the transfer function.

01:03 Now as we know transfer function of l x will be s square into s minus zero minus this will be equal to s square s.

01:17 Now we have given x zero or x dot zero will be equal to zero.

01:23 So l x will s into s minus x zero this will be s into s.

01:32 Now l of ft will be equal to f of s.

01:38 Now putting all these values in the main differential equation so we have we have m s square into s plus bs into s plus k into s that will be equal to fs.

01:59 Now this will be written as m square plus bs plus k with multiplication of s is equal to fs.

02:08 Now transfer function will be the transfer function will be one divided by m s square plus bs plus k.

02:30 Now for third part in this part we have to determine the poles and zeros of the transfer function.

02:42 Now as the transfer function is one divided by 10 square 18 0 0 0 plus 10 to the power 7.

02:50 Now we can find zeros transfer function by equating numerator equal to zero but in the given case the system has no zeros...

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