Physics- bouncing apple?

Question

An apple weighs 1.04 N. When you hang it from the end of a long spring of force constant 1.60 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.)

a) What is the unstretched length of the spring (i.e., without the apple attached)?

Answer 1

Ok, we need to remember the period for a SHM, and the period for a simple pendule :

T = 1 / F

Weight = 1.04 = m*9.8 >>> mass = 0.11 kg

F = frequency.

For the SHM, the period :

T = 2pi*sqrt( 0.11 / 1.6) = 1.65 s

then the frequency = 0.6 Hz

Now, when you change the system, and we it becomes a pendule, the frequency will decrease, and will be half of it.

New frequency = 0.3 Hz

New period = 1 / 0.3 = 3.3

For a pendule : T = 2pi*sqrt( L / g)

g = gravity = 9.8 m/s^2

3.3 = 2pi*sqrt( L / 9.8)

L = 2.7 m

Hope that helped

Answer 2

This problem is actually very funny.

Because SHM frequency is double the
side-to-side frequency there must be
very strong resonant parametric exchange
between the two modes.

No matter how small is the amplitude, the resulting
periodic change in the length of the string
(which has frequency exactly equal to that of SHM)
will rapidly grow (exponentially, I think).