When a tree branch bends, it can be assumed that acts as a spring, i.e there is a restoring force −kx in response to a vertical deflection x. Internal friction of wood provides the dominant cushioning and is independent of the hanging mass, that is, F = −bẋ with b constant. We observe that a monkey of mass m1 hangs from the branch. East monkey oscillates many times, up and down before damping off appreciably, and coming to equilibrium. Then a second monkey of mass m2 hangs from the first monkey, and we observe that the monkeys together also oscillate several times before the branch reaches equilibrium. So the new equilibrium position of the branch is three times larger than with a single monkey. Besides, the energy loss per cycle for a monkey is half that of the loss with two monkeys. Based on these observations, determine the mass of each monkey (m1 and m2) in terms of a elastic constant k of the branch and a friction coefficient of the wood b. (Note: k and b are data)
When a tree branch bends, it can be assumed that acts as a spring, i.e there is a restoring force −kx in response to a vertical deflection x. Internal friction of wood provides the dominant cushioning and is independent of the hanging mass, that is, F = −bẋ with b constant.
We observe that a monkey of mass m1 hangs from the branch. East monkey oscillates many times, up and down before damping off appreciably, and coming to equilibrium. Then a second monkey of mass m2 hangs from the first monkey, and we observe that the monkeys together also oscillate several times before the branch reaches equilibrium. So the new equilibrium position of the branch is three times larger than with a single monkey. Besides, the energy loss per cycle for a monkey is half that of the loss with two monkeys.
Based on these observations, determine the mass of each monkey (m1 and m2) in terms of a elastic constant k of the branch and a friction coefficient of the wood b. (Note: k and b are data)
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