Physics, asked by pkkp1868, 11 months ago

A particle of mass m in a unidirectional potential field have potential energy U =a+2bx^2 where a and b are constant find the time period

Answers

Answered by abhi178
14

time period of particle is 2π√{m/4b}

A particle of mass m is a unidirectional potential field have potential energy, U = a + 2bx² , where a and b are constant.

we know, force , F = -dU/dx

so, F = -d(a + 2bx²)/dx = -4bx

if particle executing simple harmonic motion, then F = -mω²x

so, F = -mω²x = -4bx

⇒ω² = 4b/m

⇒ω = √{4b/m}

we know angular frequency, ω = 2π/Time period, T

so, T = 2π/ω

= 2π/√{4b/m}

= 2π√{m/4b}

hence, time period of particle is 2π√{m/4b}

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Answered by Anonymous
3

\huge\bold\purple{Answer:-}

A particle of mass m is a unidirectional potential field have potential energy, U = a + 2bx² , where a and b are constant.

we know, force , F = -dU/dx

so, F = -d(a + 2bx²)/dx = -4bx

if particle executing simple harmonic motion, then F = -mω²x

so, F = -mω²x = -4bx

⇒ω² = 4b/m

⇒ω = √{4b/m}

we know angular frequency, ω = 2π/Time period, T

so, T = 2π/ω

= 2π/√{4b/m}

= 2π√{m/4b}

hence, time period of particle is 2π√{m/4b}

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