What is the maximum speed in a spring-mass system with low friction?

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In a horizontal spring-mass system with low friction, the maximum speed can be calculated using energy conservation principles. Given a spring stiffness of 220 N/m and a mass of 0.5 kg, the initial conditions include a spring compression of 10 cm and an initial speed of 3 m/s. The maximum speed during the motion is determined by the total mechanical energy, which remains constant barring energy dissipation. Additionally, with energy dissipation of 0.02 J per cycle, the average power input required to maintain steady oscillation can be calculated. The discussion also touches on determining the angular frequency and frequency of the oscillation.
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Homework Statement



A horizontal spring-mass system has low friction, spring stiffness 220 N/m, and mass 0.5 kg. The system is released with an initial compression of the spring of 10 cm and an initial speed of the mass of 3 m/s.

(b) What is the maximum speed during the motion?
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?

Homework Equations


The Attempt at a Solution



(a) What is the maximum stretch during the motion?
I got .1745
 
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redalertkev said:

The Attempt at a Solution



(a) What is the maximum stretch during the motion?
I got .1745

I did not check this, but I will assume it is correct.

What is the angular frequency of the oscillation and hence what is the frequency?
 
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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