A particle of mass m is moving in a potential well, for which the potential energy is given by $$U(x)=U01$$−cosax where $$U0$$ and a are constants. Then for small oscillations, time period is

Question

A particle of mass m is moving in a potential well, for which the potential energy is  given by $$U(x)=U01$$−cosax where $$U0$$  and a are constants. Then for small oscillations,  time period is

Infinity Learn · Accepted Answer

$$U(x)=U0(1$$−$$cosax)F=$$−$$dUdx=$$−$$aU0sin$$⁡ax For small oscillations, $$sin$$⁡ax≈ax⇒ma′$$=$$−$$a2U0x$$⇒a′$$=$$−$$a2U0mx$$. Since, a′α−x, oscillations are simple harmonic.  Comparing with the relation $$a=$$−ω$$2x$$ with a′$$=$$−$$a2U0mx$$ We get ω$$2=a2U0m$$. Here ω$$=2$$$$π$$T Therefore $$2$$$$π$$$$T=a2U0m$$ Hence, $$T=2$$$$π$$$$ma2U0$$.

A particle of mass m is moving in a potential well, for which the potential energy is given by $U (x) = U_{0} (1 - \cos a x)$ where $U_{0}$ and a are constants. Then for small oscillations, time period is

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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)=U01−cosax where U0 and a are constants. Then for small oscillations, time period is

Remember concepts with our Masterclasses.

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A particle of mass m is moving in a potential well, for which the potential energy is given by $U (x) = U_{0} (1 - \cos a x)$ where $U_{0}$ and a are constants. Then for small oscillations, time period is