Physics - SHM Question with Solution | TestHub

Let us displace the piston throught small distance ${\mathrm{x}}_1$ towards right,

${\mathrm{P}}_1{\mathrm{V}}_1^{\gamma }={\mathrm{P}}_2{\mathrm{V}}_2^{\gamma }={\mathrm{P}}_{\mathrm{o}}{\mathrm{V}}_{\mathrm{o}}^{\gamma }$ (Adiabatic process)

$\left(P_1-P_2\right)s=m\ddot{x}$ Now finding value of

$P_1\&P_2$. $P_1=\frac{P_o{V}_o^{\gamma }}{V_1^{\gamma }}=\frac{P_o{V}_o^{\gamma }}{{\left(V_o-xs\right)}^{\gamma }}$

$P_2=\frac{P_o{V}_o^{\gamma }}{V_2^{\gamma }}=\frac{P_o{V}_o^{\gamma }}{{\left(V_o+xs\right)}^{\gamma }}$ Now equating the equation

$P_o{V}_o^{\gamma }\left[\frac{1}{{\left(V_o-xs\right)}^{\gamma }}-\frac{1}{{\left(V_o+xs\right)}^{\gamma }}\right]S=m\ddot{x}$

$P_o{V}_o^{\gamma }\left[\frac{{\left(V_o+xs\right)}^{\gamma }-{\left(V_o-xs\right)}^{\gamma }}{{\left(V_o^2-x^2{s}^2\right)}^{\gamma }}\right]S=m\ddot{x}$

$P_o{V}_o^{\gamma }S\left[\frac{V_o^{\gamma }\left[1+\frac{xs}{V_o}\right]-V_o^{\gamma }{\left[1-\frac{xs}{V_o}\right]}^{\gamma }}{V_o^{2\gamma }{\left[1-\frac{x^2{s}^2}{V_o^2}\right]}^{\gamma }}\right]=m\ddot{x}$

Now $\frac{\gamma {\mathrm{s}}^2{\mathrm{x}}^2}{{\mathrm{V}}_{\mathrm{o}}}$ in the denominatior is very small we will neglect the term. $\frac{P_o{V}_o^{2\gamma }}{V_o^{2\gamma }}\left[1+\frac{rsx}{V_o}-1+\frac{rxs}{V_o}\right]S=m\ddot{x}$

$P_o\left[\frac{2r{S}^{2}x}{V_o}\right]=m\ddot{x}$ Now compairing the equation

$\omega =\mathrm{s}\sqrt{\frac{2{\mathrm{P}}_{\mathrm{o}}\gamma }{{\mathrm{mv}}_{\mathrm{o}}}}$z