Question

A particle of mass m and charge -q moves dimetrically through a uniform charged sphere of radius r with total charge q. the frequency of the particle simple harmonic motion if its amplitude less than r is given by

Answer 1

I don't know............. if you know than answer......

Answer 2

The frequency of the particle simple harmonic motion if its amplitude less than r is given by  $\frac{1}{2\pi } \sqrt{\frac{1}{4\pi Eo} \frac{q^{2} }{mr^{3} } }$

The General form of SHM is given by,

• $\frac{d^{2}x }{dt^{2} } + w^{2}x = 0$

Consider a uniformly charged solid sphere of radius r .

The electric field inside the sphere at a distance ' a '  is given by ,

• E = $\frac{1}{4\pi Eo}$ $\frac{Q}{r^{3} }$ a

Therefore electric field at a distance x is given by

• E =  $\frac{1}{4\pi Eo}$ $\frac{Q}{r^{3} }$ x

Let a charge -q be present at 'x '

The force on -q by the sphere is ,

• F = E( -q )

• F =  $\frac{1}{4\pi Eo}$ $\frac{Q(-q)}{r^{3} }$ x  = - $\frac{1}{4\pi Eo}$ $\frac{Qq}{r^{3} }$ x

• F = ma = m$\frac{d^{2x} }{dt^{2} }$ = - $\frac{1}{4\pi Eo}$ $\frac{Qq}{r^{3} }$ x

Comparing with the general form of SHM,

We get

• $w^{2}$ =   $\frac{1}{4\pi Eo}$ $\frac{Qq}{mr^{3} }$  ,
• Q = q (given )

Therefore frequency of oscillation, f = w /2  $\pi$

• f = $\frac{1}{2\pi } \sqrt{\frac{1}{4\pi Eo} \frac{q^{2} }{mr^{3} } }$