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?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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.

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