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Question: One end of a spring of force constant k is fixed to a vertical wall and the other to body of mass m ...

One end of a spring of force constant k is fixed to a vertical wall and the other to body of mass m resting on a smooth horizontal surface. There is another wall at a distance x0 from the body. The spring is then compressed by 2x0 and released. The time taken to strike the wall first time is –

A

B

mk\sqrt { \frac { \mathrm { m } } { \mathrm { k } } }

C

2π3 mk\frac { 2 \pi } { 3 } \sqrt { \frac { \mathrm {~m} } { \mathrm { k } } }

D

Answer

2π3 mk\frac { 2 \pi } { 3 } \sqrt { \frac { \mathrm {~m} } { \mathrm { k } } }

Explanation

Solution

The total time from A to C

tAC = tAB + tBC = + tBC

Where T = Time period of oscillation of spring-mass system tBC can be given by

BC = AB sin tBC

Putting 12\frac { 1 } { 2 }, we get

tBC =

\ tBC = 2π3 mk\frac { 2 \pi } { 3 } \sqrt { \frac { \mathrm {~m} } { \mathrm { k } } }