A uniform beam is suspended horizontally by two identical vertical springs that

Question

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 kg, and a 175-kg sack of gravel sits on the middle of it. The beam is oscillating in SHM, with an amplitude of 40.0 cm and a frequency of 0.600cycles/s. The sack of gravel falls off the beam when the beam has its maximum upward displacement. What is the frequency of the subsequent SHM of the beam? What is the amplitude of the subsequent SHM of the beam? If the gravel instead falls off when the beam has its maximum speed, what is the frequency of the subsequent SHM of the beam? What is the amplitude of the subsequent SHM of the beam?

Explanation / Answer

omega=sqrt(k/m) --> so we first find k using the initial condition, and then find the frequency after the sack falls off to be: f=0.8 Hz for both situations. For the second problem, the displacement from the equilibrium remains the same, but the actual equilibrium point changes. mg=kx is the equation we have, so solve for x to find the equilibrium positions of both situations, then you will know how far the spring is from the equilibrium point. A=9.80 cm For the third problem, we know that it has maximum velocity and so maximum kinetic energy, which is equal to 1/2k(A_old)^2. But after the sack falls off, the beam now has both kinetic energy and potential energy because the equilibrium point moves up. So you can say that the kinetic energy plus the potential energy = total energy = 1/2k(A_new)^2 We get: A=42.6 cm

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