A particle performs simple harmonic motion along a straight line with an amplitude of ten centimeters and a time period of four seconds. What is the magnitude of its acceleration at the instant when its displacement from the mean position is equal to five centimeters?

💡 Solution & Explanation

The magnitude of acceleration in simple harmonic motion is given by the product of the square of angular frequency and the displacement from the mean position. Angular frequency equals two pi divided by the time period of four seconds, yielding half pi radians per second. Squaring this angular frequency and multiplying by the five centimeter displacement yields five point zero times pi squared centimeters per second squared. Incorrect options arise from miscalculating the angular frequency or substituting the full amplitude instead of the instantaneous displacement. The restoring force in simple harmonic motion is always directed towards the equilibrium position.