A particle executes simple harmonic motion along a straight line with an amplitude of 10 centimeters and a time period of 4 seconds. What is the magnitude of its acceleration at the instant when its displacement from the mean position is 5 centimeters?

💡 Solution & Explanation

Acceleration in simple harmonic motion is given by the product of the square of angular frequency and displacement. Angular frequency equals two pi divided by the time period of 4 seconds, yielding pi over 2 radians per second. Multiplying the square of this frequency by the displacement of 5 centimeters results in 1.25 pi squared centimeters per second squared.