Measuring Blocks Equilibrium Displacements

[LCSphysicist]

Homework Statement

All below

Relevant Equations

All below

"The displacements of the blocks
from equilibrium are both measured to the right. Block 1 has a mass of 15 grams and block 2 a mass of 10 grams. The spring constants of the springs are shown in dynes/cm."
​

I don't know if i understood very well the notation, but i interpreted as F(t) acting only in x2

I started writing a system, before i go on i want to know if it is right, seems quite laborious and tiring solve it

What you think about?

## {15×(d^2 x1)/(dt^2) = -15 x1 - 90 (x1 - x2) - 15(dx1)/(dt) + cos(w t)} ##
## {(d^2 x2)/(dt^2) = -x2 - 9 (x2 - x1) - (dx2)/(dt)} ##

 

[haruspex]

Homework Statement:: All below
Relevant Equations:: All below

View attachment 266454
View attachment 266455
"The displacements of the blocks
from equilibrium are both measured to the right. Block 1 has a mass of 15 grams and block 2 a mass of 10 grams. The spring constants of the springs are shown in dynes/cm."
​

I don't know if i understood very well the notation, but i interpreted as F(t) acting only in x2

I started writing a system, before i go on i want to know if it is right, seems quite laborious and tiring solve it

What you think about?

## {15×(d^2 x1)/(dt^2) = -15 x1 - 90 (x1 - x2) - 15(dx1)/(dt) + cos(w t)} ##
## {(d^2 x2)/(dt^2) = -x2 - 9 (x2 - x1) - (dx2)/(dt)} ##

Where does the third 15 in the first equation come from? Should that be γ?
Please resist the temptation to plug in numbers straight away. Create variables as necessary and keep everything symbolic until the end. It has many advantages, including making it easier for others to follow your algebra.

 

[LCSphysicist]

Where does the third 15 in the first equation come from? Should that be γ?
Please resist the temptation to plug in numbers straight away. Create variables as necessary and keep everything symbolic until the end. It has many advantages, including making it easier for others to follow your algebra.

The third 15.
F = -bv = -m*y*v = -15*1*v

"Create variables as necessary and keep everything symbolic until the end." I totally agree with you, the problem here is that i though that maintain all the k in the solution would be a problem, because if the system is right, what immediately i would do is or assume x1 and x2 as something like A*cos + B*sin, or i would try to find a matrix what would lead me to a characteristic polynomial, so i would find w and their normal modes A (and, if this would be right, i see how tiring would be to handle the three coefficients k in a (probably) two degree equation

 

[haruspex]

F = -bv = -m*y*v = -15*1*v

Ah, yes - I forgot ##\gamma## is defined as a multiplier on the mass. Not a convention I've come across. Seems a bit weird.
Note that using symbols instead of numbers would have alerted me to that.

 

[TSny]

The differential equations of @LCSphysicist in post #1 look correct to me. I assume that the question is asking for the average rate of dissipation of energy after the transient part of the solution has died away. So, you only need to find the steady-state part of the solution (which doesn't depend on the initial conditions). It shouldn't be too bad if you use computer software to carry out the algebra.