Quick Question on mass/spring system

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The discussion revolves around calculating the total energy of a mass-spring system, where a 3 kg mass oscillates with a given amplitude and period. The spring constant was determined to be 29.6 N/m, leading to a total energy calculation of 0.333 J using the formula (.5)kA^2. Additionally, there is a query regarding the direction of a periodic wave described by the equation .15m*sin(10t+(pi)x), with the conclusion that the wave moves in a positive direction due to the positive coefficient of x. The participant seeks clarification on these concepts, confirming their understanding of both the energy calculation and wave direction.
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Homework Statement



A 3Kg mass, attached to spring oscillates w/ amplitude of .15m and a period of 2s. At time t= 0.5s the objects position is .019m.

Find the total energy of the system



Homework Equations



(.5)mv^2+(.5)kx^2= (.5)kA^2



The Attempt at a Solution



I found the spring constant to be 29.6 N/m... my only question is this... (.5)mv^2+(.5)kx^2
would give me the answer, as would (.5)kA^2 correct? to get the total energy i just did (.5)(29.6)(.15)^2 = .333J

Is that correct?
 
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Yes it is.
 
rock.freak667 said:
Yes it is.

Thank a bunch rock... i also had one other quick question that i figure ill post here so i don't start another post minutes after this one...

A periodic wave has the equation .15m*sin(10t+(pi)x)

Now i am getting a little bit mixed up about how to determine the direction of the wave... I am going to say this wave is moving in a positive direction because (pi)x is positive... meaning it is going forward... is that correct? So when i rewrite the equation it would be .15m*sin(10t+(pi)(x-vt)) to account for the positive change? Thanks for all of the help guys.
 
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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