1. ## Mass spring problem

I'm confussed to the workout of this problem.

Given a mass of 2kg, stretches a spring .392m (m i believe here is meters). At all times the damping device gives a resisting force whose magnitude in Newtons is twenty times the magnitude of the velocity in m/s. The mass is released from the equilibrium position with an upwards velocity of 5m/s. How far from the equilibrium position does the object get.

2. ## Re: Mass spring problem

Okay, what have you done on this? Can you write the equation for this motion?

3. ## Re: Mass spring problem

I'm using the equation my''+by'+ky=0. Where i'm confused is the statment of 20 times the magnitude of the velocity, which is a rate. Where do i use that rate in the equation? Where m=mass b=damping coefficient k=spring constant

4. ## Re: Mass spring problem

Try to find a solution like:

$y(\text{t})\text{=}e^{\frac{\left(-b-\sqrt{b^2-4 g k m}\right) t}{2 g m}} \text{C1}+e^{\frac{\left(-b+\sqrt{b^2-4 g k m}\right) t}{2 g m}} \text{C2}$

then find c1 & c2 based on y(0)=5....etc.

5. ## Re: Mass spring problem

Originally Posted by nivek0078
I'm using the equation my''+by'+ky=0. Where i'm confused is the statment of 20 times the magnitude of the velocity, which is a rate. Where do i use that rate in the equation? Where m=mass b=damping coefficient k=spring constant
They tell you that the damping force is "twenty times the magnitude of the velocity in m/s," which means that the value of 'b' is -20 N/(m/s). Also note that the signs of both the 'b' and the 'k' coefficients is negative (force of the spring is in a direction opposite to its displacement, and force of the damper is in a direction opposite its velocity).

6. ## Re: Mass spring problem

Can you write the equation for this motion?

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