Question

A system shown in figure, consists of a massless pulley, a spring of force constant k and a block of mass m. If block is just slightly displaced vertically down from its equilibrium position and released, then the period of vertical oscillations is–

• A. T = p $\sqrt { \left( \frac { \mathrm { m } } { 4 \mathrm { k } } \right) }$
• B. T = 2p $\sqrt { \left( \frac { \mathrm { m } } { 4 \mathrm { k } } \right) }$
• C. T =2p
• D. T = 2p$\sqrt { \left( \frac { \mathrm { m } } { 3 \mathrm { k } } \right) }$

Answer 1

Answer: (B)

T = 2p $\sqrt { \left( \frac { \mathrm { m } } { 4 \mathrm { k } } \right) }$

Answer 2

Displacement in box = x

displacement in box = 2x

F = 2 kx

̃ F¢ = 2F

F¢ = 4 kx

̃ T = 2p $\sqrt { \left( \frac { \mathrm { m } } { 4 \mathrm { k } } \right) }$