A mass of 2 kg is placed on two springs with spring constants 8 N/m and 2 N/m. What is the time period of oscillation when the mass is pressed down by 2 m.

Expert Answers
justaguide eNotes educator| Certified Educator

The frequency of oscillation of an object attached to a spring does not depend on the initial displacement. It is dependent only on the spring constant and the mass of the object. Here, the mass of the object that is placed on the springs is 2 kg.

As it is placed on both the springs, the springs are in parallel. The equivalent spring constant of two springs in parallel is the sum of their spring constants. The equivalent spring constant here is 8 + 2 = 10 N/m.

The frequency of oscillations of an object with mass m attached to a spring with spring constant k is given as F = (1/2*pi)*sqrt(k/m)

Substituting the values we have F = (1/2*pi)*sqrt(10/2)

=> (1/2*pi)*sqrt 5

The time period is the inverse of the frequency.

T = 2*pi/sqrt 5 = 2.8 sec approximately

The time period of oscillation of the block of mass 2 kg placed on springs with spring constant 8 N/m and 2 N/m is (2*pi/sqrt 5) second.