# Block Spring System

#### Razi

Please any help to solve this problem

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#### HallsofIvy

It's hard to know what in the world you meant in posting this! Surely you don't just want someone to do the problem for you since you say "help". But you show no attempt of your own to solve it so we have no way of knowing what you do know about the problem or what kind of "help" you need. We don't know your level of education- college, secondary school, elementary (unlikely I admit!). Have you studied algebra? Calculus? Differential Equations? How much physics have you studied?

I presume that you know that "force= mass x acceleration" and that the force due to a spring is "kx" where k is the "spring constant" and x is the stretch of the spring. We are told that the spring constant is 19.6 N/m. The object's mass is 1.5 kg and the applied force is 20 N so the net force is that minus the "pull back" from the spring: $20- 19.6x= 1.5\frac{d^2x}{dt^2}$. Do you know how to solve that "second order linear differential equation with constant coefficients"?

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#### Razi

Dear HallsofIvy

This question is very clear and it is found in page 415 no.21 of [ Physics for Scientists and Engineers , 5th ED ] book . My age is 44 years old and i have graduated from the university since more than 20 years ago .

I have a B.sc degree in Electrical Engineering and Diploma degree in Electronics and Diploma in Mathematics.

I know very well what is the meaning of Algebra , Calculus , Differential equations and Physics .

I know very well how to solve Differential equations
1st order , 1st degree DE , separable
Exact DE
Linear DE
singular solution - Extraneous Loci DE
1st order higher degree DE
linear DE with order n
Homogeneous Linear Equations with constant Coefficients
Linear Equations with variables Coefficients
Systems of simultaneous equations
Total DE
Partial DE
Also i know how solve all these types Numerically by hand and by Using MATLAB
Also i know how to solve some DE types by using Laplace transform with particular initial conditions

I have studied Static , Dynamic , strength of materials and thermodynamics in the 1st and 2nd year in the university , so i know very well what is the meaning of Newton's 2nd Law .

I have solved this question by using two hypothesis and i got different results so i posted this question here trying to find someone explains to me this contradiction in these two different results , this what i meant '' help'' .

Any way i have attached these two methods for solutions here and i will be very happy to find some one tell me which solution is correct .

By the way in your solution you are used Fs = Kx , but in fact it is Fs = -Kx , you are neglected the negative sign i do not know why , this negative sign guided me to the 2nd solution.

Best regards

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#### topsquark

Forum Staff
Please any help to solve this problem

Our usual approach in helping a student is to find out what the student may have done wrong or misunderstood something. Without showing your work or discussing it we don't know where to focus. HallsofIvy is simply quoting policy, which is pretty much the only way we can focus our attention on what the problem is.

Typically when I do a spring problem I don't worry so much about the negative sign. That makes writing the equations a bit more simple. Just remember that when the spring is extended the spring force is in the direction opposite the extension. (In this case that means we must use the "-" in the spring force. If the spring is compressed, then we use the "+" sign. Of course, this is for a choice of positive direction in the direction that the spring extends.) It's rather like how we choose the direction of the friction in the second part of the problem. The friction force is always acting opposite the direction of the velocity.

So anyway, the equation of motion of our object is
$$\displaystyle \dfrac{d^2x}{dt^2} = - \dfrac{k}{m}x + \dfrac{20}{m}$$
This is the equation from your first attempt.

Your solution method is a bit of overkill. Usually for spring motion we expect some kind of oscillatory motion so it's simpler to just start with the trig functions. So your homogenous solution and particular solution become
$$\displaystyle x_h(t) = A~cos( \omega t) + B~sin( \omega t)$$

$$\displaystyle x_p(t) = C$$
where $$\displaystyle \omega = \sqrt{ \dfrac{k}{m} }$$

Using x(0) = 0 and x'(0) = 0 I get
$$\displaystyle x(t) = \dfrac{20}{k} \left ( cos( \omega t) - 1 \right )$$
just as you got.

I didn't check the rest. If you need further help just let us know.

-Dan

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#### Razi

Dear Dan

Thank you so much for your clarification and for your explaining , i do appreciate that . Now , i knew that my 1st solution was correct .

Best regards

Razi