A 1.2 kg block is placed on a spring with a spring constant of 665 N/m. What would the period of oscillation of this system be? 0.0425 s 0.0425 s 0.267 s 0.267 s 798 s 798 s 554 s 554 s

Answer 1

Time period of Oscillation T = 2∏ √(m/K)

Where m = mass , K = spring constant

T= 2*3.14 ( 1.2/665) ^1/2 = 2*3.14*1.8045 * 10 ^{-3 =} 26.68*10^-2 0.2667= 0.267 Sec