Explanation:
Given that,
Spring constant of the spring, k = 6.5 N/m
Amplitude of simple harmonic motion, A = 11.5 cm
Speed of the block, v = 29 cm/s = 0.29 m/s
(a) When the block is halfway between its equilibrium position and the end point, x = 5.75 cm
The speed of oscillator at a given position is given by :



m = 0.766 kg
(b) The period of motion is given by :





T = 2.15 s
(c) The maximum acceleration of the block is given by :



Hence, this is the required solution.