- #1

14850842

- 4

- 0

Can you please help me solve the equation of motion for the following diagram

Thanks

Thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, the equation of motion for a spring-damper system is given by m * d^2x/dt^2 + b * dx/dt + kx = 0, where m is the mass of the object, b is the damping coefficient, k is the spring constant, x is the displacement of the object, and t is time. The equation can be solved using various methods, such as analytical, numerical, or graphical methods, with the Laplace transform method being a common approach. The initial conditions required for solving the equation are the initial displacement x(0) and initial velocity dx/dt(0) at time t = 0. The damping coefficient affects the motion by determining the amount of resistance,

- #1

14850842

- 4

- 0

Can you please help me solve the equation of motion for the following diagram

Thanks

Thanks

Physics news on Phys.org

- #2

Thaakisfox

- 263

- 0

- #3

14850842

- 4

- 0

What I need is the equation, the rest I can work from there

- #4

- #5

blue_raver22

- 2,250

- 0

In this system, we have a mass (m) attached to a spring with a spring constant (k) and a damper with a damping coefficient (c). The displacement of the mass from its equilibrium position is represented by x(t).

To solve the equation of motion, we can use Newton's second law, which states that the sum of forces acting on an object is equal to its mass multiplied by its acceleration. In this case, the forces acting on the mass are the spring force (F_s) and the damping force (F_d).

F_s = -kx(t)

F_d = -c(dx/dt)

Therefore, the equation of motion for this system can be written as:

m(d^2x/dt^2) + c(dx/dt) + kx = 0

This is a second-order differential equation, and to solve it, we need to apply initial conditions. These conditions can be the initial position (x_0) and velocity (v_0) of the mass at time t=0.

Using these initial conditions, we can solve the equation of motion using mathematical techniques such as Laplace transforms, or numerical methods such as Euler's method or Runge-Kutta method.

I hope this helps you in solving the equation of motion for your system. Let me know if you have any further questions or need clarification on any of the concepts mentioned.

The equation of motion for a spring-damper system is given by *m * d^2x/dt^2 + b * dx/dt + kx = 0*, where *m* is the mass of the object, *b* is the damping coefficient, *k* is the spring constant, *x* is the displacement of the object from its equilibrium position, and *t* is time.

The equation of motion for a spring-damper system can be solved using various methods, such as analytical, numerical, or graphical methods. One common approach is to use the Laplace transform method, where the equation is transformed into the s-domain and then solved for *x(s)*. The inverse Laplace transform is then applied to obtain the solution in the time domain.

The initial conditions required to solve the equation of motion for a spring-damper system are the initial displacement *x(0)* and initial velocity *dx/dt(0)* of the object at time *t = 0*. These initial conditions are necessary to obtain the complete solution of the equation.

The damping coefficient *b* determines the amount of resistance to the motion of the object. A higher damping coefficient results in greater resistance and therefore, a slower decay in amplitude. On the other hand, a lower damping coefficient allows for more oscillations before the amplitude decays to zero.

The spring constant *k* determines the stiffness of the spring in the system. A higher spring constant results in a stiffer spring, which leads to a greater force resisting the displacement of the object. This results in a shorter period of oscillation and a quicker decay in amplitude. A lower spring constant, on the other hand, leads to a longer period of oscillation and a slower decay in amplitude.

- Replies
- 16

- Views
- 3K

Automotive
Spring and damper on rocker

- Replies
- 2

- Views
- 1K

Engineering
Control System mass spring damper

- Replies
- 0

- Views
- 1K

- Replies
- 1

- Views
- 2K

- Replies
- 1

- Views
- 8K

- Replies
- 15

- Views
- 6K

- Replies
- 3

- Views
- 2K

- Replies
- 9

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Replies
- 16

- Views
- 3K

Share: