A simple pendulum of length `L ` is suspended from the top of a flat beam of thickness `L//2` . The bob is pulled away from the beam so that it makes an angle `thetalt30^(@)` with the vertical , as shown in figure. It is then released from rest. If `phi` is the maximum angular deflection to teh right, then

Options

A. `phi=theta`
B. `phi ge 2theta`
C. `phi lt theta`
D. `thetaltphilt2theta`

Answer

D

Explanation

Refer to figure , At `A`
`KE=0, PE=mg(L-Lcostheta)`
`=mgL(1-costheta)`
Total energy at `A` is
`E_(A)=0+mgL(1-costheta)`
`mgL(1-costheta)`
At `B`, `KE=0`
`PE=mg[(L)/(2)-(L)/(2) cosphi]=(mg)(L)/(2)[1-cosphi]`
Total energy at ` B` is
`E_(B)=0+(mg)(L)/(2)[1-cosphi](mg)(L)/(2)[1-cosphi]`
Accoding to law of conservation of energy
`E_(A)-E_(B)`
`mgL(1-costheta)=(mg)(L)/(2)(1-cosphi)`
or `1-cos theta =(1-cosphi)/(2)`
or `2-2costheta=1-cosphi`
or `2costheta=1+cosphi` or `cos theta=(1+cosphi)/(2)`
`:. theta lt phi lt 2theta`
Hence, option (d ) is correct.