Question

A ring of radius 8 cm that lies in the ya plane carries positive charge of 2 μC uniformly distributed over its length. A particle of mass m that carries a charge of-2 oscillates about the center of the ring with an angular frequency of 22 rad/s Find the angular frequency of oscillation of the mass if the radius of the ring is doubled

Answer 1

The angular frequency of oscillation of the mass is 11 rad/s.

Step-by-step explanation:

Given that,

Charge = 2μC

Radius R₁= 8 cm

Radius R₂ = 16 cm

Angular frequency = 22 rad/s

We need to calculate the angular frequency of oscillation of the mass

The electric field produced along x axis

`E=(kqx)/(√(R^2+x^2))`

`E=\frac{kqx}{\sqrt{R^2(1+(x^2)/(R^2))^2}}`

`E=(kqx)/(R^3)`

The force on the mass is

`F=Eq`

`F=(kQqx)/(R^3)`....(I)

For,x<<R

Now, using centripetal force

`F = (mv^2)/(r)`

Put the value of F in equation (I)

`(mv^2)/(r)=(kQq)/(R^3)`

We know that,

`v=r\omega`

`m\omega^2r=(kQq)/(R^3)`

`\omega^2=(kQq)/(mrR^3)`

For, r<<R

`\omega^2=(kQq)/(mR^3)`

`\omega=\sqrt{(kQq)/(mR^3)}`

Here,

`\omega\propto\sqrt{(q)/(R^3)}`

The ratio of angular frequency

`(\omega)/(\omega_(1))=\sqrt{((q)/(R^3))/((q_(1))/(R_(1)^3))}`

`(\omega)/(\omega')=\sqrt{(R^3*2q)/(2R^3* q)}`

`\omega=\sqrt{(8^3*2*2*10^(-6))/(8*8^3*2*10^(-6))}*\omega'`

`\omega=0.5\omega'`

Put the value of

`\omega=(1)/(2)*22`

`\omega=11\ rad/s`

Hence, The angular frequency of oscillation of the mass is 11 rad/s.