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| 1. |
A particle executes S.H.M. with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of tis velocity is equalto that of its acceleration. Then its time periodin second is : |
| Answer» <html><body><p>`(1)/(2pi sqrt(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))`<br/>`2pi sqrt(3)`<br/>`(2pi)/(sqrt(3))`<br/>`(sqrt(3))/(2pi)`</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> :`omega sqrt(r^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)-y^(2))=omega^(2)y` <br/> `sqrt(r^(2)-y^(2))=omega y=(2pi)/(T)y` <br/> `T=(2pi y)/(sqrt(r^(2)-y^(2)))=(2pi <a href="https://interviewquestions.tuteehub.com/tag/xx1-1463705" style="font-weight:bold;" target="_blank" title="Click to know more about XX1">XX1</a>)/(sqrt(2^(2)-1^(2)))=(2pi)/(sqrt(3))`second. <br/> Correct choice is ( c ).</body></html> | |