A particle of mass `m` moving along x-axis has a potential energy `U(x)=a+bx^2` where a and b are positive constant. It will execute simple harmonic motion with a frequency determined by the value of

A particle of mass m moves in a one dimensional potential energy U(x)=-ax^2+bx^4 , where a and b are positive constant. The angular frequency of small oscillation about the minima of the potential energy is equal to

A particle of mass m moves in a one dimensional potential energy U(x)=-ax^2+bx^4 , where a and b are positive constant. The angular frequency of small oscillation about the minima of the potential energy is equal to

A particle of mass m moves in a one dimensional potential energy U(x)=-ax^2+bx^4 , where a and b are positive constant. The angular frequency of small oscillation about the minima of the potential energy is equal to

If a particle of mass m moves in a potential energy field U=U_(0)-ax+bx^(2) where U_(0) , a and b are positive constants.Then natural frequency of small oscillations of this particle about stable equilibrium point is (1)/(x pi)sqrt((b)/(m)) . The value of x is?

A particle free to move along x-axis is acted upon by a force F=-ax+bx^(2) whrte a and b are positive constants. , the correct variation of potential energy function U(x) is best represented by.