24u can get some hint from here
http://targetiit.com/iit_jee_forum/posts/find_min_friction_1375.html[1].....though that ques is diff but concept is same
the angular amplitudeof a simple pendulum is (theta)0 The maximum tension in the string will be    a)mg(1+(theta)o2)         b)mg(1-(theta)o2)          c)mg (1+(theta)o)             d)mg(1-(theta)o)
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2 Answers
24
3okay so when the thing comes to mean position...........
mg(L-Lcos(θ)) = 1/2mv2 ......
2gL(1-cos(θ)) = v2 ...........
tension at mean position =
mg + mv2/L .......
mg + 2mg(1-cos(θ)) .. .........(1)
mg + 4mgsin2(θ/4) .....
mg + 4mgθ2/4 .......[for small θ] ...
mg(1+ θ2).................
[actully i think theta shud be half of angular amplitude....................
otherwise in eq. 1 ill have to put θ/2 which dosnt bring the desired answer][1]
- Anonymous How " mg + 4mg sin square theta/4" come from "mg + 2mg (1 -cos theta)"Upvote·0·2018-05-21 21:39:09