A movable pulley of negligible mass carries a block of mass m. One end of an inextensible string is fixed to the ceiling; the string passes under the movable pulley and its other end is attached to the lower end of a vertical spring of force constant k, whose upper end is fixed. For a small downward displacement y of the movable pulley from equilibrium, the change in spring extension is 2y. The angular frequency of small vertical oscillations of the block is:
Options

Question

A movable pulley of negligible mass carries a block of mass m. One end of an inextensible string is fixed to the ceiling; the string passes under the movable pulley and its other end is attached to the lower end of a vertical spring of force constant k, whose upper end is fixed. For a small downward displacement y of the movable pulley from equilibrium, the change in spring extension is 2y. The angular frequency of small vertical oscillations of the block is:

Options

G1 Physics · Accepted Answer

A downward displacement y of the movable pulley changes the spring extension by 2y, so the change in string tension is ΔT = k(2y)=2ky. The movable pulley is supported by two string segments, hence the change in total upward force is 2ΔT=4ky. Thus the restoring equation is mÿ = -4ky, giving ω = √(4k/m)=2√(k/m).