Question

A spring of spring constant 7600N/m is attached to a block of mass 0.25 kg. frequency of oscillation on the frictionless surface is

Answer 1

Explanation:

When displaced from equilibrium, the net force exerted by the springs is  –2kx  acting in a direction so as to return the block to its equilibrium position  (x=0). Since the acceleration  a=d2x/dt2, Newton’s second law yields,  mdt2d2x=−2kx. 

Substituting  x=xmcos(ωt+ϕ)  and simplifying, we find  ω2=m2k

where  ω  is in radians per unit time. Since there are  2π  radians in a cycle, and frequency  f  measures cycles per second, we obtain

f=2πω=2π1m2k=2π10.245kg2(7580N/m)=39.6Hz.

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