Mass attached to a horizontal spring

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A particle with a mass of 0.210 kg is attached to a horizontal spring with a force constant of 0.84 N/m and has a maximum speed of 5 m/s moving left at t=0. The equation of motion for the particle is derived using the spring force and mass acceleration equations. Initially, there was confusion in isolating the position variable x, but the correct solution was found. The final equation of motion is expressed as x(t) = (5/2)Sin(2t + π). This demonstrates the oscillatory motion of the mass-spring system.
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A particle with a mass of 0.210kg is attached to a horizontal spring with a force constant of 0.84 N/m. At the moment t=0, the particle has its maximum speed of 5m/s and is moving to the left. (Assume that the positive direction is to the right.)

Determine the particle's equation of motion, specifying its position as a function of time (use the following as necessary: t)

F = -kx
F = ma



What I did was:
-0.84x = 0.210a
-0.84x = 0.210[(v - 5)/t)]
-0.84x = 0.210[(x/t - 5)/t]


I still can't isolate for x, and i don't know if what i am doing is right. Please help.
 
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nvm, i figured it out, its 5/2Sin(2t+pi)
 
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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