- #1

bolzano95

- 89

- 7

- Homework Statement
- There is an mass-spring oscillator made of a spring with stiffness k and a block of mass m. The block is affected by a friction.

At the time ##t=0s## the block is pulled from the equilibrium position for ##x=5cm## in the right direction and released.

Derive and solve the equation for a displacement x, velocity v and power P.

- Relevant Equations
- II. Newton's Law

There is an mass-spring oscillator made of a spring with stiffness

$$F_f = -k_f N tanh(\frac{v}{v_c})$$

##k_f## - friction coefficient

N - normal force

##v_c## - velocity tolerance.

At the time ##t=0s## the block is pulled from the equilibrium position for ##x=5cm## in the right direction and released.

1. Derive the differential equation.

2. Solve it for a displacement x.

3. Solve it for a velocity v.

4. Solve it for the power loss P, because of the friction.

1. Deriving the necessary equation:

In the vertical direction: N=mg.

Because the only forces acting on the block in horizontal direction are the spring force and friction we can write:

$$-kx + k_f\, N\, tanh(\frac{\dot{x}}{v_c})=m\ddot{x}$$

$$m\ddot{x} - k_f\: mg\: tanh(\frac{\dot{x}}{v_c}) +kx=0$$

Here my solving stops, because I'm not sure how to implement the standard solution ##x(t)= Ce^{\lambda t}## because velocity is inside the function ##tanh(\frac{\dot{x}}{v_c})##.

**k**and a block of mass**m**. The block is affected by a friction given by the equation:$$F_f = -k_f N tanh(\frac{v}{v_c})$$

##k_f## - friction coefficient

N - normal force

##v_c## - velocity tolerance.

At the time ##t=0s## the block is pulled from the equilibrium position for ##x=5cm## in the right direction and released.

1. Derive the differential equation.

2. Solve it for a displacement x.

3. Solve it for a velocity v.

4. Solve it for the power loss P, because of the friction.

1. Deriving the necessary equation:

In the vertical direction: N=mg.

Because the only forces acting on the block in horizontal direction are the spring force and friction we can write:

$$-kx + k_f\, N\, tanh(\frac{\dot{x}}{v_c})=m\ddot{x}$$

$$m\ddot{x} - k_f\: mg\: tanh(\frac{\dot{x}}{v_c}) +kx=0$$

Here my solving stops, because I'm not sure how to implement the standard solution ##x(t)= Ce^{\lambda t}## because velocity is inside the function ##tanh(\frac{\dot{x}}{v_c})##.