Help Solve Diff EQ Problem: Vibrating Spring w/ 64 lb Weight

In summary, the problem involves finding the function y, which represents the position of a weight attached to a spring, as a function of time given certain initial conditions and external forces. The equation of motion is determined by the weight, gravity, damping factor, spring constant, and the external force. The solution to the homogeneous equation is used to find a particular solution, and then the initial conditions are applied to find the final solution. However, in this particular problem, the driving frequency is the same as the natural frequency, making the solution more complex and unstable.
  • #1
ptyindian
1
0
Hi i need some help with a differential equations problem. The question is:-
(w/g)y(double prime) + ky(single prime)+cy=F(t)
w= the weight of the object attached to the spring
g=32
k= damping factor
c=spring constant
F(t)= the external force on the system.
A 64 pound weight is attached to the end of the spring. After reaching the equilibrium position the spring is stretched one foot beyond the equilibrium position. The weight is then released and as it is released it is struck a downward blow giving it an initial velocity of 2 ft/sec. Take the moment the weight is released and struck as time zero. At time zero a periodic external force given by F(t) = (1/2)cos(4t) pounds begins acting on the system. t is time in seconds. Consider the damping factor to be negligible, i.e., take k to be zero. The spring constant is 32. Find the function giving y, the position of the bottom of the weight as a function of time given in seconds.
if some one could help me with this problem i would really appreciate it.
Thanks
 
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  • #2
ptyindian said:
Hi i need some help with a differential equations problem. The question is:-
(w/g)y(double prime) + ky(single prime)+cy=F(t)
w= the weight of the object attached to the spring
g=32
k= damping factor
c=spring constant
F(t)= the external force on the system.
A 64 pound weight is attached to the end of the spring. After reaching the equilibrium position the spring is stretched one foot beyond the equilibrium position. The weight is then released and as it is released it is struck a downward blow giving it an initial velocity of 2 ft/sec. Take the moment the weight is released and struck as time zero. At time zero a periodic external force given by F(t) = (1/2)cos(4t) pounds begins acting on the system. t is time in seconds. Consider the damping factor to be negligible, i.e., take k to be zero. The spring constant is 32. Find the function giving y, the position of the bottom of the weight as a function of time given in seconds.
if some one could help me with this problem i would really appreciate it.
Thanks

first of all I am somewhat confused in terms of the notation
typically c is the damping constant and k is the spring constant
I will use this notation in my discussion here...

the EOM can be written as

y'' + (k/m)*y = F/m*cos(wd*t)

here k is the spring constant, m is the mass (w/g), and wd is the driving frequency.

Notice that the system has two complex roots with entirely imaginary parts. The solution to the homogeneous equation...

y'' + (k/m)*y = 0 takes the form

y = c1*cos(w*t) + c2*sin(w*t)

where c1 and c2 are arbitrary constants that are determined by the initial conditions and w is the natural frequency of system which is given by (k/m)^1/2. However, we cannot apply the boundary conditions yet to solve for c1 and c2. We must first consider the driving term.

Notice that for this problem w = sqrt(k/m) = sqrt(k*g/w) = sqrt(16) = 4
and wd = 4. This is not good, this means the system is going to be highly unstable because the driving frequency is the same as the natural frequency. Thus the amplification factor is infinit. So by using the method of undetermined coefficients we need to look for a solution of the form
A*cos(wd*t) + B*sin(wd*t) however this is not entirely valid because the new particular solution is not independent of the homogeneous solution. Thus we are really looking for a new solution of the form y = t*(A*cos(w*t)+B*sin(w*t)).

Now we have to solve

y'' + 16*y = 1/4*cos(4*t)

use y = t*(A*cos(w*t)+B*sin(w*t)).
and differentiate twice to find and expression for y''

Now solve for A and B such that the coefficient of the sin term is zero
and the coefficient of the cosine term is 1/4

You will have 2 eqns and 2 unknowns such that

F(A,B)*cos(4*t) + G(A,B)*sin(4*t) = 1/4*cos(4*t)

So solve F(A,B) = 1/4 and G(A,B) = 0
this will give you the particular solution.

The entire solution has the form y = c1*cos(4*t) + c2*sin(4*t) + Yp

where Yp = t*(A*cos(4*t)+B*sin(4*t)) where you have solved for A and B above

Now with A and B known plug into

y = c1*cos(4*t) + c2*sin(4*t) + Yp

and now apply the initial (boundary conditions) which are the initial displacement and initial velocity.

notice that c1*cos(4*t) + c2*sin(4*t) can be written as R*sin(4*t+p)

where R = sqrt(c1^2+c2^2) and p = atan(c1/c2)

In the end you should have a sin wave for 1 term and then another sne wave with a secular term (term that grows with time)

I would expect a solution of the form

(C+D*t)*sin(4*t + E)

where C D and E are numbers.

This problem is significantly ugly because the driving and natural frequencies are the same. Essentially this is like a person pushing you on a swing at the same frequency at which the swing would naturally osciallate. If we neglectthe air resistance and friction in the links this basically states that after enough pushes you would end up going in circles around the pole holding the swing!
 

1. What is a vibrating spring with a 64 lb weight?

A vibrating spring with a 64 lb weight refers to a physical system in which a spring is attached to a 64 lb weight and is free to oscillate or vibrate.

2. What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. In this case, the differential equation for a vibrating spring with a 64 lb weight would describe the relationship between the displacement of the spring and its velocity and acceleration.

3. How can I solve a differential equation for a vibrating spring with a 64 lb weight?

To solve a differential equation for a vibrating spring with a 64 lb weight, you will need to use mathematical techniques such as separation of variables or Laplace transforms. You will also need to apply specific initial conditions to obtain a unique solution.

4. What factors affect the frequency of a vibrating spring with a 64 lb weight?

The frequency of a vibrating spring with a 64 lb weight is affected by several factors, including the stiffness of the spring, the mass of the weight, and the initial conditions of the system. A higher stiffness or a larger mass will result in a higher frequency, while different initial conditions can result in different frequencies.

5. Why is it important to understand and solve differential equations for vibrating springs with a 64 lb weight?

Understanding and solving differential equations for vibrating springs with a 64 lb weight is important for several reasons. Firstly, it allows us to model and predict the behavior of physical systems, which is crucial in many scientific and engineering fields. Additionally, it helps us understand the underlying principles and concepts of differential equations, which are widely used in various areas of mathematics and science.

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