Extremely quick question (MV calc/waves)

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The discussion centers on the confusion regarding a negative sign in the partial derivative of a wave function in a physics context. The original poster questions whether the negative sign is correct after taking the derivative. Responses clarify that the negative sign is indeed appropriate, arising from the derivative of the cosine function and the wave equation's structure. Emphasis is placed on understanding the fundamentals to build confidence in solving such problems. Overall, the conversation encourages a solid grasp of the concepts to avoid uncertainty in physics calculations.
Zorodius
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My physics book gives this function for the displacement of a wave element oscillating in a longitudinal wave. Later on, they take the partial derivative with respect to time, like so.

My question: Did my brain fall out, or is the negative sign in the second picture not supposed to be there? I'm open to either possibility right now :biggrin:
 
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You are correct. You get one negative from the kx-\omega t and another from the derivative of cosine.
 
Zorodius said:
My question: Did my brain fall out, or is the negative sign in the second picture not supposed to be there? I'm open to either possibility right now :biggrin:

Don't be. Understand your basics well enough that you know when something's right and when it's wrong. Your brain is doing good...just needs to feel more confident, and confidence comes from familiarity. :smile:
 
Thanks for your replies!
 
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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