A pendulum bob of mass `m` hangs form a massless elastic string. The potential energy (elastic + gravitational) of the system (bob + string + earth) measured relative to the position of the bob corresponding to the normal length of the string is: (where `x` = static deformation (elongation) of the string.)

Options

A. `mgx`
B. `-(1)/(2)mgx`
C. `2mgx`
D. `-mgx`

Answer

B

Explanation

Potential energy of bob `w.r.t` given refference `= -mgx`
elastic potential energy of spring `= (1)/(2) kx^(2)`
`= (1)/(2) (mg)/(x) xx x^(2) = (1)/(2) mgx`
Hence total potential energy of the system
`= -mgx + (1)/(2) mgx = - (1)/(2) mgx`.