Solving Algebra: How do we go from 1 to 2?

Thread starter jimmy42
Start date Mar 13, 2012
Tags

Algebra

AI Thread Summary

The discussion centers on a query about the transition from the first equation, 1 -x(t)= −Bω2 cos(ωt) −Cω2 sin(ωt), to the second equation, 2 - x(t)=−ω2 x(t). A participant suggests that the correct interpretation involves rewriting the first equation to show that 0 x(t) equals Bcos(ωt) + Csin(ωt). They clarify that the second equation represents the second derivative of x(t), denoted as x''(t), which equals -ω²x(t). The conversation highlights the importance of understanding the relationship between derivatives and the functions involved in algebraic expressions.

Mar 13, 2012

jimmy42

Hello,

I am having some problem with some algebra:

1 -x(t)= −Bω2 cos(ωt) −Cω2 sin(ωt)
2 - x(t)=−ω2 x(t)

Can someone explain how we went from 1 to 2?

Thanks.

Physics news on Phys.org

Mar 13, 2012

tiny-tim

Science Advisor

Homework Helper

hello jimmy!

i'll guess that that should read …

0 x(t)= Bcos(ωt) + Csin(ωt)

1 x''(t)= -Bω²cos(ωt) - Cω²sin(ωt)

2 x''(t) = −ω²x(t)

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »