Laplace tx word problem

[tremor]

Hi!
I have an optional HW problem that I would like to get a handle on in case something like it ends up on an exam. Any help would be greatly appreciated!

A gun on a tank is attached to a spring-mass-dashpot system with spring constant of 100?^2 and a damping constant of 200?. The mass of the gun is 100. Assume that the displacement of the gun from its rest position after being fired at t=0 is y(t). The equation describing y(t) is:

100(d^2 y)/(dt^2 )+200? dy/dt+100?^2 y=0

and y(0) = 0; dy(0)/dt = 100

It is desired that one second after firing, the quantity y^2+ (dy/dt)^2 should be less than 0.01. How large must ? be to guarantee this will be so?

I've tried to solve. I'm not sure if it is correct though, especially with the ? term still in there.

Y(s) = s^2/(?^2(s+1)^2)=A/?^2(s+1)^2 + B/?^2(s+1)

= A?^2(s+1) + B?^2(s+1)^2*(s+1)/(?^2(s+1)^2)

Does this Laplace transform seem right? I'm still new to rearranging the equations. Thanks!!

 

[tremor]

Word problem aside, could someone help me with the algebra here I know it isn't that difficult but it just hasn't clicked how to do this yet:

I know I have an equation that looks like this:
100(s^2Y(s)-sy(0)-y'(0))+200?(sY(s)-y(0))+100V^2(Y(s))= 0

Y(s)=-sy(0)+y'(0)+200?(y(0)/100?^2(s^2 + ??) this is where I get a little lost - I know I'm just rearranging to solve for Y(s) but I'm not entirely sure how this turns out. Could someone help me just figure out the correct denominator? Thanks!

 

[galactus]

Hi!
I have an optional HW problem that I would like to get a handle on in case something like it ends up on an exam. Any help would be greatly appreciated!

A gun on a tank is attached to a spring-mass-dashpot system with spring constant of 100?^2 and a damping constant of 200?. The mass of the gun is 100. Assume that the displacement of the gun from its rest position after being fired at t=0 is y(t). The equation describing y(t) is:

\(\displaystyle 100(d^2 y)/(dt^2 )+200{\alpha} dy/dt+100{\alpha}^{2} y=0\)

and \(\displaystyle y(0) = 0; dy(0)/dt = 100\)

Since \(\displaystyle y''=p^{2}Y-py_{0}-y'_{0}, \;\ y'=pY-y_{0}, \;\ y=Y\)

so, we get: \(\displaystyle 100\left[p^{2}Y-100\right]+200{\alpha}\cdot pY+100{\alpha}^{2}Y=0\)

\(\displaystyle 100p^{2}Y-10000+200{\alpha}pY+100{\alpha}^{2}Y=0\)

Factor out Y:

\(\displaystyle Y\left(100p^{2}+200{\alpha}p+100{\alpha}^{2}\right)=10000\)

\(\displaystyle Y=\frac{10000}{100p^{2}+200{\alpha}p+100{\alpha}^{2}}\Rightarrow \frac{100}{{\alpha}^{2}+2{\alpha}p+p^{2}}=\frac{100}{({\alpha}+p)^{2}}\)

Look this up in a LaPlace table and we see it is \(\displaystyle \boxed{y=100te^{-at}}\)

It is desired that one second after firing, the quantity y^2+ (dy/dt)^2 should be less than 0.01. How large must ? be to guarantee this will be so?

\(\displaystyle y^{2}+(\frac{dy}{dt})^{2}=\left((1000a^{2}+10000)t^{2}-20000at+10000\right)e^{-at}<\frac{1}{100}\)

If t=1, then we get \(\displaystyle (10000a^{2}-20000a+20000)e^{-2a}<\frac{1}{100}\)

\(\displaystyle (a^{2}-2a+2)e^{-2a}<\frac{1}{1000000}\)

Solve for a. Use some sort of software if you have a rough time doing it algebraically.

 

[tremor]

thank you so much!! that helped clear up a lot. i really appreciate it!

 

[chad146]

Hi!
I have an optional HW problem that I would like to get a handle on in case something like it ends up on an exam. Any help would be greatly appreciated!

A gun on a tank is attached to a spring-mass-dashpot system with spring constant of 100?^2 and a damping constant of 200?. The mass of the gun is 100. Assume that the displacement of the gun from its rest position after being fired at t=0 is y(t). The equation describing y(t) is:

\(\displaystyle 100(d^2 y)/(dt^2 )+200{\alpha} dy/dt+100{\alpha}^{2} y=0\)

and \(\displaystyle y(0) = 0; dy(0)/dt = 100\)

Since \(\displaystyle y''=p^{2}Y-py_{0}-y'_{0}, \;\ y'=pY-y_{0}, \;\ y=Y\)

so, we get: \(\displaystyle 100\left[p^{2}Y-100\right]+200{\alpha}\cdot pY+100{\alpha}^{2}Y=0\)

\(\displaystyle 100p^{2}Y-10000+200{\alpha}pY+100{\alpha}^{2}Y=0\)

Factor out Y:

\(\displaystyle Y\left(100p^{2}+200{\alpha}p+100{\alpha}^{2}\right)=10000\)

\(\displaystyle Y=\frac{10000}{100p^{2}+200{\alpha}p+100{\alpha}^{2}}\Rightarrow \frac{100}{{\alpha}^{2}+2{\alpha}p+p^{2}}=\frac{100}{({\alpha}+p)^{2}}\)

Look this up in a LaPlace table and we see it is \(\displaystyle \boxed{y=100te^{-at}}\)

It is desired that one second after firing, the quantity y^2+ (dy/dt)^2 should be less than 0.01. How large must ? be to guarantee this will be so?

\(\displaystyle y^{2}+(\frac{dy}{dt})^{2}=\left((1000a^{2}+10000)t^{2}-20000at+10000\right)e^{-at}<\frac{1}{100}\)

If t=1, then we get \(\displaystyle (10000a^{2}-20000a+20000)e^{-2a}<\frac{1}{100}\)

\(\displaystyle (a^{2}-2a+2)e^{-2a}<\frac{1}{1000000}\)

Solve for a. Use some sort of software if you have a rough time doing it algebraically.

Greetings,

Your answer makes no sense to me. What is \(\displaystyle y''=p^{2}Y-py_{0}-y'_{0}, \;\ y'=pY-y_{0}, \;\ y=Y\) ??
How you are going from y'' to y=Y --> what is Y in this case?
Thanks.

 

[Deleted member 4993]

Greetings,

Your answer makes no sense to me. What is \(\displaystyle y''=p^{2}Y-py_{0}-y'_{0}, \;\ y'=pY-y_{0}, \;\ y=Y\) ??
How you are going from y'' to y=Y --> what is Y in this case?
Thanks.

Have you studied Laplace Transformation - at all?