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PhysicsSHMCalculation of time periodMedium2 minQB

PhysicsMediummultiple choice

A particle of mass m has potential energy given by

Options:(select one or more)

Answer:

B, C

Solution:

$F = - \frac{d U}{d x} = - U_{0} [- \frac{4 x}{x _{0}^{2}} + \frac{4 x ^{3}}{x _{0}^{4}}] = 0 \Rightarrow x = 0, \pm x_{0}$ .

$\frac{d ^{2} U}{d x ^{2}} = U_{0} [- \frac{4}{x _{0}^{2}} + \frac{12 x ^{2}}{x _{0}^{4}}]$ : positive at $x = \pm x_{0}$ (stable - B), negative at $x = 0$ (unstable).

$ω = \frac{U ^{''} ( x _{0} )}{m} = \frac{8 U _{0}}{m x _{0}^{2}}$ - (C)

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: QB

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