A 0.45-kg crow lands on a slender branch and bobs up and down with a period of 1

Question

A 0.45-kg crow lands on a slender branch and bobs up and down with a period of 1.1 s. An eagle flies up to the same branch, scaring the crow away, and lands. The eagle now bobs up and down with a period of 3.6 s. Treating the branch as an ideal spring, find:
(a) the effective force constant of the branch and
(b) the mass of the eagle.

Explanation / Answer

Let the spring constant be K. Now, For a simple harmonic motion, T= 2pv(m/K) Where m is the mass of the spring. This equation is derived by solving the equation of shm x(t)=acos (?t+f) Solving this we get, accn =a =?2x = ALso, ma= force = -kx... since it is a spring. =>?2=k/m We know t=2p/? =2pv(0.45/K) =>(1.1/2p)2= 0.45/K =>K=0.150 N/m Again.. Let mass of crow is M SO, using the same formula, t=2pv(M/K) =>(3.6/2p)2=M/K =>M=4.712810-3 Kg =4.7 gm

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