The Shortest distance travelled by a particle executing SHM from mean position in 2 s is equal to sqrt(3)//2 times its amplitude . Determine its time period.

The Shortest distance travelled by a particle executing SHM from mean

1.	The Shortest distance travelled by a particle executing SHM from mean position in 2 s is equal to sqrt(3)//2 times its amplitude . Determine its time period.
Answer» <html><body><p><br/></p>Solution :`y = a <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> omegat , (<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(3//<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))a = sin omegat, sin omegat = sqrt(3//2)= sinpi//3, t = 2 s, (2pi//T) = pi//3 , T= <a href="https://interviewquestions.tuteehub.com/tag/12s-271682" style="font-weight:bold;" target="_blank" title="Click to know more about 12S">12S</a>`</body></html>