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The potential energy of a particle of mass 1 kg in motion along x-axis is given by U=(10 - 10 cos6x)J. the period of small oscillation is π^2 approximate 10.

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Given potential energy of oscillating mass at displacement x is

U(x)=10-10cos6x

So periodic force (F(x)) acting on the oscillating mass of 1kg   will be

F(x)= -dU/dx= -10×6sin6x

For small oscillation sin6x =6x

So F(x)= -360x

So periodic acceleration

a(x)=F(x)/1=360x

Now by definition of SHM we know that periodic acceleration a(x) on the oscillating mass is related with w the angular frequency of the reference point associated with SHM

a(x)=-w^2x,  

So w^2 = 360

=>w= 6√10

=>4π^2/T^2=360

=>T^2=π^2/90

=>T^2=10/90=1/9 (inserting π^2=10)

=>T=1/3 s

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