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the angular amplitudeof a simple pendulum is (theta)₀ The maximum tension in the string will beÂ Â Â Â a)mg(1+(theta)_o²)Â Â Â Â Â Â Â Â Â b)mg(1-(theta)_o²)Â Â Â Â Â Â Â Â Â Â c)mg (1+(theta)_o)Â Â Â Â Â Â Â Â Â Â Â Â Â d)mg(1-(theta)_o)

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eureka123 ·2009-02-26 19:13:26

u 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

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iitimcomin ·2009-02-26 19:46:27

okay so when the thing comes to mean position...........

mg(L-Lcos(Î¸)) = 1/2mv² ......

2gL(1-cos(Î¸)) = v² ...........

tension at mean position =

mg + mv²/L .......

mg + 2mg(1-cos(Î¸)) .. .........(1)
mg + 4mgsin²(Î¸/4) .....

mg + 4mgÎ¸²/4 .......[for small Î¸] ...

mg(1+ Î¸²).................

[actully i think theta shud be half of angular amplitude....................
otherwise in eq. 1 ill have to put Î¸/2 which dosnt bring the desired answer][1]

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Anonymous How " mg + 4mg sin square theta/4" come from "mg + 2mg (1 -cos theta)"
Upvote·0·2018-05-21 21:39:09
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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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shm

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