The bob of a simple pendulum of length L = 2 m has a mass m = 2kg and a speed u=1 ms−1  when the string is at 370 to the vertical. (cos 370=0.8)

Speed of the particle at the lowest point is first calculated by the law of conservation of energy, so(U+K)A=(U+K)B mgL(1−cosθ)+12mu2=0+12mv2⇒ v2=u2+2gL(1−cosθ)         ……..(1)Now, tension at the lowest point is     T=mL(vB2+gL), where vB=v ⇒ T=mL(u2+3gL−2gLcosθ)Now, according to question,u=1 ms−1,g=10  ms−2,L=2m, m=2kg⇒  T=(22)(1+60−32)   ⇒ T=29 N (b) At the highest point, we haveNow, please do not confuse θmax with θ as both may be different. So let us first calculate θmax. For this purpose we shall try to calculate the total mechanical energy of the particle at A, which isEA=UA+KA         ⇒  EA=mgL(1−cosθ)+12mu2⇒ EA=(2)(10)(2)(1−45)+12(2)(1)2   ⇒EA=8+1=9 JSo, now at maximum height, we have vc=0 and by Law of conservation fo energy, total energy at C should also be 9 J⇒ mgL(1−cosθmax)=9   ⇒  (2)(10)(2)(1−cosθmax)=9⇒ 1−cosθmax=940          ⇒ cosθmax=3140⇒  TC=(2)(10)(3140)        ⇒  TC=15.5N