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

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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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