Question

# A smooth wedge of mass m and angle of inclination 60 degrees rests unattached between two springs of spring constant k and 4k, on a smooth horizontal plane, both springs in the un-extended position. The time period of small oscillations of the wedge (assuming that the springs are constrained to get compressed along their length) equals

A

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B
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C

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D

None of above

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Solution

## The correct option is C If wedge moved left by x spring will be compressed by y Horizontal force on wedge = Fx =- ky sin 60∘ = ma = −4k × (sin 60)2=ma = −4k×34=ma a=−3km x T=2πω=2π√m3k t1=T2=π√m3k Since the wedge is in contact with the spring for half the time period Similarly for the right spring Time period will be T=2π√mk t2=T2=π√mk total tiem of oscillation= t1+t1=π√mk(1+1√3) option B is correct

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