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
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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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}