A 1 kg block is placed on a frictionless surface with springs attached to opposite sides with a spring constant 8 and 24 N/m. What is the frequency of oscillation?

Expert Answers
justaguide eNotes educator| Certified Educator

Two springs are attached in opposite directions of the block. The springs are attached to the block in parallel. The fact that they are in opposite directions does not make a difference as the resistive force created when a spring is stretched or compressed is the same, only in the direction is reversed in either case.

The equivalent resistance of the springs is Ke where 1/Ke = 1/K1 + 1/K2. Here, the spring constant of the springs is 8 and 24. Ke = (8*24)/(8 + 24) = 8*24/32 = 6 N/m

The frequency of oscillation of a harmonic oscillator created with a spring of constant k and mass m is f = (1/2*pi)*sqrt(k/m)

This gives the required frequency of oscillation as f = (1/2*pi)*sqrt 6

The frequency of oscillation of the block attached to the springs is sqrt 6/2*pi Hz.