A simple pendulm has a length L and a bob of mass M. The bob is vibrating with amplitude a .What is the maximum tension in the string?

Options

A. `T_("max")=mg[1+((A)/(l))^(2)]`
B. `T_("max")=mg[1-((A)/(l))^(2)]`
C. `T_("max")=mg[1-((l)/(A))^(2)]`
D. `T_("max")=mg[1+((l)/(A))^(2)]`

Answer

A

Explanation

The tension in the string is maximum, when the bob is at the equilibrium position O. It is moving in a circle of radius r = l.
`T_("max")=mg+(mv^(2))/(r )`
When it is at C, its total energy `(E_(C ))`= mgh and when it is at O its total energy `(E_(O))=K.E. =(1)/(2) mv^(2)`
`because E_(C )=E_(O) therefore mgh =(1)/(2)mv^(2) therefore v^(2)=2gh`
U"sin"g (2) in (1), we get
`T_("max")=mg +(m(2gh))/(l) (because r =l)`
`=mg (1+(2h)/(l))`
From the figure,
`h= l-l "cos" theta =l (1 -"cos" theta) =l . 2 "sin"^(2)(theta)/(2)`
`because angle =("arc")/("radius")`
`therefore theta =(OC)/(l)=(A)/(l) (because` arc = amplitude A)
`therefore (theta)/(2)=(A)/(2l)`
and for small angles, `"sin" theta= theta`
`therefore "sin"^(2)((theta)/(2))= ((A)/(2l))^(2)=(A^(2))/(4l^(2))`
`therefore " from " (4), h=l . 2 (A^(2))/(4l^(2))=(A^(2))/(2l)`
`therefore (2h)/(l)=(2A^(2))/(2l. l)=(A^(2))/(l^(2))=((A)/(l))^(2)`
U"sin"g (6) in (3), we get
`T_("max")=mg [1+((A)/(l))^(2)]`
This is option (a).