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A mass of 3 kg stretches a spring 9m. The mass is acted on by an external force of 2 AND. The Mass moves in a medium that imparts a viscous force of 1 N when the speed of the mass is 4m/sec The mass is pulled down 8 cm below its equilibrium position, and then set in motion inthe upward direction with a velocity of 5 m/sec. State the initial value problem describing the motion of the mass. DO NOT SOLVE.

Sagot :

Answer:

  k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

give us some initial conditions

1) friction force fr = 1N when v = 4m / s

2) an initial displacement of x = 0.08 m for t=0 s

Explanation:

In this exercise, you are asked to state the problem you are posing. We are going to find the equation of motion for this exercise. Let's start with Newton's second law

Let's set a reference system with the y-axis in a vertical and positive direction upwards.

We have four forces: an external downward force, negative in sign, the but that goes down and is negative, the Hook force that goes up and is positive and the friction force that opposes the movement, in this case it goes down being negative

let's write Newton's second law

          F_e -F -fr - W = m a

where

          F_e = -kDy = - k y

          fr = - b v = -b dy / dt

          W = mg

we substitute for the specific case, that is, using the signs

          k y  -b [tex]\frac{dy}{dt}[/tex] - m g - F = m [tex]\frac{d^2y}{dt^2}[/tex]

In the initial condition of the problem, before starting the movement, the friction force is zero and the acceleration is also zero

         k y - m g - F = 0

from this equation you can find the spring constant, y= 9m and F=2 N

It is not clear if when the movement starts this external force becomes zero, but since it balances the weight we can eliminate the two forces that have the same magnitude and opposite direction, so the equation remains

              k y - b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

give us some initial conditions

1) friction force fr = 1N when v = 4m / s

2) an initial displacement of x = 0.08 m for t=0 s

therefore, to initiate the movement, a small external force F 'is applied that moves the system to a new equilibrium position and this small force F' is made zero, thus initiating an oscillatory movement, described by the equation.

             k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

This is a differential equation of the second degree, therefore it needs two initial conditions for its complete solution

The initial amount of displacement corresponds to the amplitude of movement A = 0.08 m

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