The bob of a pendulum has its rest point `1` metre below the support. The bob is pulled aside until the string makes an angle of `15^(@)` with the vertical. Upon release, with what speed does the bob swing past its rest point ?

Explanation

Here, `l=1m, theta=15^(@)`
`K.E.` at rest point `= P.E.` from the starting point
`(1)/(2)m upsilon^(2)=mgl(1-costheta)`
`upsilon=sqrt(2gl(1-costheta)) = sqrt(2xx10xx1(1-cos15^(@)))`
`=sqrt(20(1-0.9659))`
`upsilon=sqrt(20xx0.0341)=0.83m//s`