Suspended from a vertical spring, a block of mass 0.5 kg undergoes simple harmonic motion with an amplitude of 0.1 m and a time period of 0.314 s . Determine the maximum force exerted by the spring on the block.

Step-by-step solution

Angular frequency = 2 T = 2 3.14 0.314 = 20 rad/s Maximum acceleration of the block during SHM is a_ max = ^2 A = (20)^2 0.1 = 40 m/s ^2 The maximum force exerted by the spring occurs at the lowest point of the oscillation where the spring is maximally stretched and the acceleration is directed upwards. Using Newton's second law at the lowest point: F_ max - mg = m a_ max F_ max = m(g + a_ max ) Taking g = 10 m/s ^2 : F_ max = 0.5 (10 + 40) = 0.5 50 = 25 N