What went wrong with my solution for a damped mass spring system?

In summary: Gerbils are inferior.In summary, the conversation revolves around solving the equation x'' + 10x' + 64x = 0 with initial conditions x(0) = 1 and x'(0) = 0. The general solution is given as e^(-5t)(c1*cos(6.245t) + c2sin(6.245t)), but the final solution is incorrect. After some recalculations and solving for the unknown coefficient c2, the correct solution is obtained.
  • #1
FHamster
8
0
So the equation is x'' + 10x' + 64x = 0
x(0) = 1
x'(0) = 0
I get general solution of e^(-5t)(c1*cos(6.245t) + c2sin(6.245t) )
From there I get e^(-5t)cos(6.245t)+5e^(-5t)sin(6.245t)
but it wrong. What the gerbils am I doing wrong?

Thanks
 
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  • #2
Welcome to PF!

Hi FHamster! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)
FHamster said:
So the equation is x'' + 10x' + 64x = 0
x(0) = 1
x'(0) = 0

From there I get e^(-5t)cos(6.245t)+5e^(-5t)sin(6.245t)
but it wrong. What the gerbils am I doing wrong?

(do hamsters not like gerbils? :biggrin:)

erm :redface: … haven't you left out a 6.245 ? :wink:
 
  • #3
FHamster said:
So the equation is x'' + 10x' + 64x = 0
x(0) = 1
x'(0) = 0
I get general solution of e^(-5t)(c1*cos(6.245t) + c2sin(6.245t) )
From there I get e^(-5t)cos(6.245t)+5e^(-5t)sin(6.245t)
but it wrong. What the gerbils am I doing wrong?

Thanks
If [itex]y= e^{5t}(c_1cos(6.245t)+ c_2sin(6.245t))[/itex]
then [itex]y(0)= e^0 (c_1 cos(0)+ c_2 sin(0))= c_1= 1[/itex] so you have that coefficient right.

[tex]y'= 5e^{5t}(c_1 cos(6.245t)+ c_2 sin(6.245t))+ e^{5t}(-6.245 sin(6.245 t)+6.245cos(6.245t))[/tex]

[tex]y'(0)= 5e^0(c_1 cos(0)+ c_2 sin(0))+ e^0(-6.245c_1 sin(0)+ 6.245c_2 cos(0))[/tex]
[tex]y'(0)= 5(c_1)+ (1)(6.245c_2)= 11.245[/tex]
Knowing that [itex]c_1= 1[/itex], solve for [itex]c_2[/itex].
 
  • #4


Thanks, following some of these guidelines and doing some recalcumacations. I managed to get it right

tiny-tim said:
Hi FHamster! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)


(do hamsters not like gerbils? :biggrin:)

erm :redface: … haven't you left out a 6.245 ? :wink:


Hamsters are the masterrace.
 
  • #5
for sharing your work and question! It seems like you are on the right track with finding the general solution for this damped mass spring system. However, there may be a mistake in your final solution. The general solution you provided is correct, but the specific solution that satisfies the initial conditions of x(0)=1 and x'(0)=0 is actually e^(-5t)cos(6.245t). This can be found by plugging in the initial conditions and solving for the constants c1 and c2. So the final solution should be e^(-5t)cos(6.245t). It is possible that you may have made a mistake when solving for the constants or when plugging in the initial conditions. I would recommend double checking your work and calculations to see if you can identify where the mistake may have been made. Keep up the good work!
 

1. What is a damped mass spring system?

A damped mass spring system is a physical system composed of a mass connected to a spring and a damper. The mass is free to move vertically, and the spring provides a restoring force while the damper dissipates energy and prevents the system from oscillating indefinitely.

2. How does damping affect the behavior of a mass spring system?

Damping affects the behavior of a mass spring system by reducing the amplitude of the oscillations and slowing down the system's response to external forces. It also decreases the system's natural frequency and changes the shape of its oscillations.

3. What are the different types of damping in a mass spring system?

There are three types of damping in a mass spring system: underdamping, critical damping, and overdamping. Underdamping occurs when the damping coefficient is less than the critical damping coefficient, resulting in oscillations with decreasing amplitude. Critical damping occurs when the damping coefficient is equal to the critical damping coefficient, and the system returns to equilibrium without overshooting. Overdamping occurs when the damping coefficient is greater than the critical damping coefficient, causing the system to return to equilibrium slowly without oscillating.

4. How is the damping coefficient calculated in a mass spring system?

The damping coefficient in a mass spring system can be calculated using the formula c = 2mωζ, where m is the mass, ω is the natural frequency, and ζ is the damping ratio. The damping ratio is determined by the type of damping, with underdamping having a ζ value less than 1, critical damping having a ζ value of 1, and overdamping having a ζ value greater than 1.

5. What real-life applications use damped mass spring systems?

Damped mass spring systems have many real-life applications, such as shock absorbers in cars, suspension systems in buildings, and vibration control systems in machines. They are also used in musical instruments, such as pianos and guitars, to produce sustained and controlled vibrations.

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