Question

the maximum velocity of a particle executing simple harmonic motion is 100 cm per second and the maximum acceleration is 157 cm per second square determine the periodic time

Answer 1

The time period of the particle executing SHM will be 0.25 sec

Explanation :

For a particle executing simple harmonic motion we know that ,

Vmax = Aω

and

a max = Aω²

where A is the amplitude of SHM and

ω is the angular velocity,

dividing both we get,

a max/Vmax = Aω²/Aω

=> 157/100 = ω

=> ω = 1.57

=> 2πT = 1.54

=> 2 x 3.14 x T = 1.54

=> T = 1/4 = 0.25 sec

Hence the time period of the particle executing SHM will be 0.25 sec