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A particle is executing a simple harmonic motion. Its maximum acceleration is alpha and maximum velocity is beta. Then, its time period of vibration will be……… |
| Answer» <html><body><p>`(2pi <a href="https://interviewquestions.tuteehub.com/tag/beta-2557" style="font-weight:bold;" target="_blank" title="Click to know more about BETA">BETA</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a>)`<br/>`(beta^2)/(alpha^2)`<br/>`(alpha)/(beta)`<br/>`(beta^2)/(alpha)`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a> acceleration in <a href="https://interviewquestions.tuteehub.com/tag/shm-630759" style="font-weight:bold;" target="_blank" title="Click to know more about SHM">SHM</a> <br/> `a_("max")= A omega^(2)` <br/> `therefore alpha = A omega^(2)` and <br/> Maximum velocity `v_("max")= A omega` <br/> `therefore beta = A omega` <br/> `therefore (a_("max"))/(v_("max"))= (A omega^(2))/(A omega)` <br/> `(alpha)/(beta)= omega` <br/> `therefore (alpha)/(beta)= (2pi)/(T)` <br/> `therefore T= 2pi (beta)/(alpha)`.</body></html> | |