Problem 12 An apple weighs 1.18 N . When you hang it from the end of a long spri
ID: 1482861 • Letter: P
Question
Problem 12
An apple weighs 1.18 N . When you hang it from the end of a long spring of force constant 1.51 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.)
Part A
What is the unstretched length of the spring (i.e., without the apple attached)?
Problem 12
An apple weighs 1.18 N . When you hang it from the end of a long spring of force constant 1.51 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.)
Part A
What is the unstretched length of the spring (i.e., without the apple attached)?
Explanation / Answer
(1/(2pi))*sqrt(1.51/(1.18/9.8)) = .54665 = bounce frequency
so side to side frequency is one half this or = .273325
0.2733 = (1/(2pi))*sqrt(g/L)
solving for L i get L = 3.33 m
Then to find the L of the unstreched spring i used F=-kx
Since the gravity exerts a downward force on the apple, its force will be negative.
So you'll get :
-1.18 = -1.51*(3.33-x), which gives you x = 2.5526 m
Calculations maybe faulty,logic is correct.
Hope this helps :)