Acceleration of a particle a= -bx, where x is the displacement of particle from mean position and b is constant. What is the periodic time of oscillation ?

Acceleration of a particle a= -bx, where x is the displacement of part

1.	Acceleration of a particle a= -bx, where x is the displacement of particle from mean position and b is constant. What is the periodic time of oscillation ?
Answer» <html><body><p>`2pi <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(b)`<br/>`(2pi)/(sqrt(b))`<br/>`(2pi)/(b)`<br/>`2 sqrt(((pi)/(b))`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/comparing-2531631" style="font-weight:bold;" target="_blank" title="Click to know more about COMPARING">COMPARING</a> `a= -kx` with `a= -omega^(2)x, omega^(2)= b` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> omega = sqrt(b)` <br/> `therefore (2pi)/(T)= sqrt(b)` <br/> `therefore T= (2pi)/(sqrt(b))`.</body></html>