A particle executing simple harmonic motion starts from mean position with amplitude ' A ' and periodic time ' T '. At what displacement is its speed one-fourth of the maximum speed?

Step-by-step solution

The velocity of a particle executing simple harmonic motion at a displacement x from the mean position is given by v = A^2 - x^2 . The maximum velocity of the particle is v_ max = A . Given that v = v_ max 4 . Substituting the expressions for v and v_ max gives A^2 - x^2 = A 4 . Squaring both sides yields A^2 - x^2 = A^2 16 . Rearranging the terms gives x^2 = A^2 - A^2 16 = 15A^2 16 . Taking the square root gives x = A 15 4 . Answer: A 15 4