A simple pendulum has time period T_(1). The point of suspension is now moved upwards according to the relation =kt^(2),Ik=1m//sec^(2)) where y is the vertical displacement. The time period now becomes T_(2) then find the ratio of (T_(1)^(2))/(T_(2)^(2)) (g=10m//sec^(2))

A simple pendulum has time period T_(1). The point of suspension is no

1.	A simple pendulum has time period T_(1). The point of suspension is now moved upwards according to the relation =kt^(2),Ik=1m//sec^(2)) where y is the vertical displacement. The time period now becomes T_(2) then find the ratio of (T_(1)^(2))/(T_(2)^(2)) (g=10m//sec^(2))
Answer» <html><body><p></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> :`y-kt^(2)/<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>//2at^(2)` <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> <br/> `implies1/2a=k=1""impliesa=2m//sec^(2)` <br/> `T_(1)=2pisqrt(l/<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)` and `T_(2)=2pisqrt(l/(g+a))` <br/> `(T_(1)^(2))/(T_(2)^(2))=(g+a)/g=(10+2)/10=6/5`</body></html>