quiz24_solutions

# quiz24_solutions - of the oscillation ANS Since the masses...

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05/09/08 Physics 201 Quiz 24 Consider an Atwood machine, masses m1, m2, with a massless pulley. The pulley (radius R) cannot, however, rotate freely; it has a piece of wire running through its center that acts as a torsional spring, with tau=-k theta. Calculate the frequency of the Atwood oscillator: ANS the kinetic energy is 1/2(m1+m2)R^2 (d(theta)/dt)^2, and the spring potential energy is 1/2 k theta^2. So the oscillation frequency is f=omega/(2 pi)=1/(2 pi)sqrt(k/ ((m1+m2)R^2)) The two masses are held at equal initial heights, and then released. What is the amplitude
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Unformatted text preview: of the oscillation? ANS Since the masses are not moving initially, the energy of the system is completely potential energy. If we find the equilibrium position of the oscillator, this will be where the potential energy is zero and the kinetic energy is maximum. Thus the difference between the initial position and the equilibrium position is the amplitude of the oscillation. tau=-k theta+(m1-m2)gR. Thus in equilibrium theta=(m1-m2)gR/k, the magnitude of which is the amplitude...
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