A simple pendulum is vibrating with an angular amplitude of 90^(@) as shown in the figure. For what value of a is the acceleration directed i) vertically upwards ii) horizontally iii) vertically downward

A simple pendulum is vibrating with an angular amplitude of 90^(@) as

1.	A simple pendulum is vibrating with an angular amplitude of 90^(@) as shown in the figure. For what value of a is the acceleration directed i) vertically upwards ii) horizontally iii) vertically downward
Answer» <html><body><p>`0^(@), cos^(-1)(1/sqrt3), 90^(@)`<br/>`90^(@), cos^(-1)(1/sqrt3), 0^(@)`<br/>` cos^(-1)(1/sqrt3),0^(@), 90^(@)`<br/>` cos^(-1)(1/sqrt3),90^(@), 0^(@)`</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> :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_NEET_PHY_XI_P2_C06_E11_006_S01.png" width="80%"/> <br/> At "C", `T cos alpha = <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> .......(i)` <br/>`P.E_(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) = K.E_(A) Rightarrow mgr = 1/2(mv_(A)^(2))` <br/> hence at "C" , `v = <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a> (v_(A)^(2) - 2gr(1-cos alpha))` <br/> `T -Mg cos alpha = (Mv)^(2)/(r) = (M.2 <a href="https://interviewquestions.tuteehub.com/tag/gr-1010452" style="font-weight:bold;" target="_blank" title="Click to know more about GR">GR</a> cos alpha)/(r)`</body></html>