A particle is performing simple harmonic motion about x = 0 with an amplitude ' a ' and periodic time T. The speed of the particle at x = a 3 will be

Step-by-step solution

The velocity of a particle in simple harmonic motion at a distance x from the mean position is given by v = a^2 - x^2 where is the angular frequency and a is the amplitude. Given x = a 3 and = 2 T , substituting these values gives v = 2 T a^2 - ( a 3 )^2 v = 2 T a^2 - a^2 9 v = 2 T 8a^2 9 v = 2 T ( 2 2 a 3 ) v = 4 2 a 3T Answer: 4 2 , a 3T