An apple weighs 1.00 N . When you hang it from the end of a long spring of force

Question

An apple weighs 1.00 N . When you hang it from the end of a long spring of force constant 1.44 N/m and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back and forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (i.e., without the apple attached)?

Explanation / Answer

mass of apple=weight/g=1.04/9.8=0.1061 kg when apple bounces vertically, the frequency is given by f1=2*pi*sqrt(k/m) where k=force constant m=mass of the apple f1=2*pi*sqrt(1.54/0.1061)=23.937 Hz when it swings side by side, the frequency is given by 2*pi*sqrt(l/g) where l=unstretched length f2=2*pi*sqrt(l/9.8) given that f2=0.5*f1 solving for l,we get l=35.568 m

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