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If a particle of mass m moves in a poten...

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?

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Knowledge Check

  • 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
    `pisqrt((a)/(2b))`
    B
    `2sqrt((a)/(m))`
    C
    `sqrt((2a)/(m))`
    D
    `sqrt((a)/(2m))`

  • A particle of mass m in a unidirectional potential field have potential energy U (x) = a + 2b x^(2) where a and b are positive constants. Find time period of oscillation :-

    A
    `2pi sqrt((2b)/(m))`
    B
    `2pi sqrt((m)/(2b))`
    C
    `pi sqrt((m)/(b))`
    D
    `pi sqrt((b)/(m))`

  • A particle of mass m is moving in a potential well, for which the potential energy is given by U(x) = U_(0)(1-cosax) where U_(0) and a are positive constants. Then (for the small value of x)

    A
    the time period of small osciallation is `T=2pisqrt(m/(aU_(0))`
    B
    the speed of the particle is maximum at x=0
    C
    the amplitude of oscillations is `pi/8`
    D
    the time period of small osciallations is `T=2pisqrt(m/(a^(2)U_(0))`

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