What does wt stand for in oscillation equations

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In oscillation equations, "wt" represents angular displacement, where "w" is the angular frequency calculated as w=2π/T and "t" is time. The discussion highlights the confusion around how "wt" relates to position, velocity, and acceleration in these equations. It clarifies that the maximum values of sine and cosine functions are 1, which means the maximum values of position and velocity can be represented as A and B, respectively. Understanding this concept helps in deriving the maximum speed and acceleration from the equations. Overall, grasping the role of "wt" is crucial for solving related homework problems effectively.
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What does "wt" stand for in oscillation equations

Hi,

In my physics class we are on the oscillation chapter. I'm kind of confused as to what "wt" in the position/velocity/accel. equations actually is. I understand that w=2pi/T and t is the time in an equation x=Acos(wt + delta) but when asked to find the max speed and accelerations after derving their formuls, the trig "disapprears" (assuming delta=o.)

What does "wt" stand for together and how does this change depending if the acceleration is max or the speed is max?

(I know this isn't a specific problem but it's a concept that I see on some of my HW probs.)
 
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'wt' is essentially angular displacement.

But when it comes to trig functions if y=Bsinx or Bcosx

the maximum value of both sine and cosine is 1, so that the maximum value of 'y' is B.
 


OH! I see. Thanks a lot. :)
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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