A particle of mass 10kg is placed in potential field given by #V = (50x^2 + 100)# erg/g. What will be frequency of oscillation of particle ?

1 Answer
Jane
Jan 28, 2018

For a conservative force acting, relation between potential energy and force is given as #F= -(dU)/(dx)# (here, given,#V#= potential ,so #V*m#=#U=#potential energy=#10000(50x^2+100)#,see the unit)

So, differentiating the given equation,we get,

#F= -100*10000x# or, #F/m =-100*10000/10000x# = #-100 x#

Now, in case of a particle in S.H.M force equation is given as, #F/m= - omega^2x#

So,comparing the obtained force equation,we get, #omega =sqrt(100)# i.e #10 (rad)/s#

This is the angular frequency.

So, #nu# = #omega/(2 pi) = 1.6 s^-1#

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