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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

π(1+12)mk

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B
π(1+13)mk
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C

π(1+13)mk

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D
None of above
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Solution

The correct option is B π(1+13)mk

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=2πm3k

t1=T2=πm3k

Since the wedge is in contact with the spring for half the time period

Similarly for the right spring

Time period will be

T=2πmk

t2=T2=πmk

total tiem of oscillation= t1+t1=πmk(1+13)

option B is correct


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