A) \[\frac{1}{2\pi }\frac{{{A}_{\gamma }}{{P}_{0}}}{{{V}_{0}}M}\]
B) \[\frac{1}{2\pi }\frac{{{V}_{0}}M{{P}_{0}}}{{{A}^{2}}\gamma }\]
C) \[\frac{1}{2\pi }\sqrt{\frac{{{A}^{2}}\gamma {{P}_{0}}}{M{{V}_{0}}}}\]
D) \[\frac{1}{2\pi }\sqrt{\frac{M{{V}_{0}}}{{{A}_{\gamma }}{{P}_{0}}}}\]
Correct Answer: C
Solution :
As adiabatic process \[P{{V}^{\gamma }}=\]Constant \[\frac{dp}{p}+\gamma \frac{dv}{v}=0\] \[dp=-\frac{p\gamma }{V}dv=-\frac{P\gamma }{V}(A)x\] (\[x\]is small displacement) \[F=(dp)A=-\frac{{{P}_{0}}\gamma {{A}^{2}}}{{{V}_{0}}}x\] \[a=-\frac{{{P}_{0}}\gamma {{A}^{2}}}{m{{v}_{0}}}x\] \[\omega =\sqrt{\frac{{{P}_{0}}\gamma {{A}^{2}}}{m{{v}_{0}}}}\] \[f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{{{P}_{0}}\gamma {{A}^{2}}}{m{{v}_{0}}}}\]
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