A particle of mass m is moving in a potential well, for which the potential energy is given by U(x)=U0(1-cos⁡ax) where U0 and a are constants. Then time period of small oscillations is:

F = -dU/dx = -U0 a sin⁡axFor small oscillations, sin⁡ax = ax⇒F = -U0 a2 x⇒accelerationnA=Fm=−U0a2mx⇒ω=U0a2m⇒T=2πmU0a2

A particle of mass m is moving in a potential well, for which the potential energy is given by U(x)=U0(1-cos⁡ax) where U0 and a are constants. Then time period of small oscillations is:

F = -dU/dx = -U0 a sin⁡axFor small oscillations, sin⁡ax = ax⇒F = -U0 a2 x⇒accelerationnA=Fm=−U0a2mx⇒ω=U0a2m⇒T=2πmU0a2

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A particle of mass m is moving in a potential well, for which the potential energy is given by U(x)=U₀(1-cos⁡ax) where U₀ and a are constants. Then time period of small oscillations is:

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$T = 2 π \sqrt{\frac{m}{a^{2} U_{0}}}$

$T = 2 π \sqrt{\frac{m}{{aU}_{0}}}$

particle does not oscillate

$T = π \sqrt{\frac{m}{U_{0}}}$

answer is A.

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Detailed Solution

F = -dU/dx = -U₀ a sin⁡ax
For small oscillations, sin⁡ax = ax⇒F = -U₀ a² x
$\Rightarrow$ acceleration $nA = \frac{F}{m} = - \frac{U_{0} a^{2}}{m} x \Rightarrow ω = \sqrt{\frac{U_{0} a^{2}}{m}} \Rightarrow T = 2 π \sqrt{\frac{m}{U_{0} a^{2}}}$