The effective length of a simple pendulum is 1 m and the mass of its bob is 5 g. If the amplitude of motion of the pendulum is 4 cm, what is the maximum tension on the string to which the bob is attached?

Explanation

The velocity of the bob is maximum at its position of equlibrium. The maximum velocity `v_(max)=omegaA=(2piA)/T`.
At this position, centripetal force is also maximum, whose value is `F_(c)=(mv_(max)^(2))/L=(momega^(2)A^(2))/L`
`" "` [L = effective length of the pendulum]
The resultant of the weight mg of the bob and tension F on the string becomes equal to this force `F_(c)`.
`therefore" "F=mg+F_(c)=mg+(momega^(2)A^(2))/L=m(g+(omega^(2)A^(2))/L)`
Again, `T=2pisqrt(L/g)`, so, `omega=(2pi)/T=sqrt(g/L)or,omega^(2)=g/L`
`therefore" "F=m(g+(gA^(2))/L^(2))=mg(1+A^(2)/L^(2))`
`=5xx980(1+4^(2)/100^(2))" "[becauseL=1m=100cm]`
`=4908"dyn"=0.04908N~~0.05N`