Tessellate - S.T.E.M.S - Physics - College - Subjective Problem

Consider a thin ring of radius a and uniform charge density λ \lambda . A particle of mass m and charge -q is placed near the center of the ring on the axis of symmetry and released. The particle will oscillate along the axis of symmetry. Find the frequency of oscillation.

Note by Writabrata Bhattacharya
3 years, 5 months ago

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Comments

The differential vertical acceleration of the particle is day=−kqλam(x2+a2)(xx2+a2)dθd{ a }_{ y }=-\frac { kq\lambda a }{ m({ x }^{ 2 }+{ a }^{ 2 }) } (\frac { x }{ \sqrt { { x }^{ 2 }+{ a }^{ 2 } } } )\quad d\theta , where kk is the Coulomb's constant, xx is the distance of the particle from the center of the ring, and dθd\theta is an angle subtended by a segment of the ring with differential length. Hence, ay=−kqλam(x2+a2)3/2x∫02πdθ=−2πkqλam(x2+a2)3/2x=−2πkqλma2((x/a)2+1)3/2x=−2πkqλma2x(1−32(x/a)2+...)≈−2πkqλma2x { a }_{ y }=-\frac { kq\lambda a }{ m{ ({ x }^{ 2 }+{ a }^{ 2 }) }^{ 3/2 } } x\int _{ 0 }^{ 2\pi }{ d\theta } =-\frac { 2\pi kq\lambda a }{ m{ ({ x }^{ 2 }+{ a }^{ 2 }) }^{ 3/2 } } x=-\frac { 2\pi kq\lambda }{ m{ a }^{ 2 }{ ({ (x/a) }^{ 2 }+1) }^{ 3/2 } } x=-\frac { 2\pi kq\lambda }{ m{ a }^{ 2 } } x\quad (1-\frac { 3 }{ 2 } { (x/a) }^{ 2 }+...)\approx -\frac { 2\pi kq\lambda }{ m{ a }^{ 2 } } x\quad . Therefore ω=2πkqλma2\omega =\sqrt { \frac { 2\pi kq\lambda }{ m{ a }^{ 2 } } } .

Rui-Xian Siew - 3 years, 5 months ago
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