A0.45-kg crow lands on a slender branch and bobs up and down with aperiod of1.1

Question

A0.45-kg crow lands on a slender branch and bobs up and down with aperiod

of1.1 s. An eagle flies up to the same branch, scaring the crow away,and lands. The eagle

nowbobs up and down with a period of 3.6 s. Treating the branch as anideal spring, find (a)

theeffective force constant of the branch and (b) themass of the eagle.

Explanation / Answer

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

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