A particle of mass 10g lies in a potential field U=(50`x^(2)`+50)erg/g.The value of frequency of oscillation in hertz will be "

A particle of mass 10gm is placed in a potential field given by V = (50x^(2) + 100)J//kg . The frequency of oscilltion in cycle//sec is

A particle of mass 4 gm. Lies in a potential field given by V = 200x^(2) + 150 ergs/gm. Deduce the frequency of vibration.

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 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

Figure. Shows the variation of force acting on a particle of mass 400 g executing simple harmonic motion. The frequency of oscillation of the particle is

Figure. Shows the variation of force acting on a particle of mass 400 g executing simple harmonic motion. The frequency of oscillation of the particle is

The variation of force (F) acting on a particle of mass 800 g with position (x) is as shown in figure. The time period of oscillation of the particles is

A particle of mass 10g is undergoing SHM of amplitude 10cm and period 0.1s. The maximum value of force on particle is about

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 sinusoidal wave travels along a taut string of linear mass density 0.1g//cm . The particles oscillate along y- direction and wave moves in the positive x- direction . The amplitude and frequency of oscillation are 2mm and 50Hz respectively. The minimum distance between two particles oscillating in the same phase is 4m . The tension in the string is ( in newton )