A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.12 m The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of block?

Solution 23E Step 1: Mass of the block = m in Kg Amplitude A = 0.12 m Maximum speed v max 3.90 m/s Step 2: To find the magnitude of maximum acceleration Maximum acceleration a max = kA m Where k is spring constant. Mass and spring constant (force constant) are not given directly. We need to calculate the ratio of mass and spring constant to find maximum acceleration. Step 3: From conservation of energy 1mv +xkx 1 2 = kA 2 2 2 2 Velocity of the particle is maximum when it is in the mean position where x = 0 therefore 1 2 1 2 2mv max = k2 v = k A or max m