A classical mechanics problem by Rishabh Deep Singh

F.B.D.

$If\quad the\quad plank\quad is\quad displaced\quad by\quad x\quad towards\quad right\quad then.\\ Let\quad { N }_{ 1 },\quad { N }_{ 2 }\quad and\quad { f }_{ 1 }\quad ,\quad { f }_{ 2 }\quad are\quad normal\quad and\quad friction\\ Forces\quad at\quad Points\quad A\quad and\quad B.\\ By\quad force\quad balancing\quad \\ { N }_{ 1 }+{ N }_{ 2 }=Mg\quad --------(1)\\ By\quad torque\quad balancing\quad about\quad point\quad A\\ Mg(L+x)={ N }_{ 2 }.(2L)\quad \quad \quad \quad --------(2)\\ Using\quad equation\quad (1)\quad and\quad (2)\\ { N }_{ 1 }=\frac { Mg }{ 2 } -\frac { Mgx }{ 2L } \quad \quad and\quad { N }_{ 2 }=\frac { Mg }{ 2 } +\frac { Mgx }{ 2L } \\ So\quad { f }_{ 1 }{ =\mu N }_{ 1 }=\mu \left( \frac { Mg }{ 2 } -\frac { Mgx }{ 2L } \right) \\ and{ \quad f }_{ 2 }{ =\mu N }_{ 2 }=\mu \left( \frac { Mg }{ 2 } +\frac { Mgx }{ 2L } \right) \\$

F.B.D. of M

${ f }_{ 2 }{ -f }_{ 1 }=Ma\\ \mu \left( \frac { 2Mgx }{ 2l } \right) =Ma=-\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } .M\\ \frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } =-\frac { \mu gx }{ l } \\ Comparing\quad with\\ \frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } =-{ \left( \omega \right) }^{ 2 }.x\\ \omega =\sqrt { \frac { \mu g }{ l } } \\ Time\quad period\quad T=\frac { 2\pi }{ \omega } \\ T=2\pi \sqrt { \frac { l }{ \mu g } } \\ Frequency(\nu )=\frac { 1 }{ 2\pi } \sqrt { \frac { \mu g }{ l } } \\ \nu =\frac { 1 }{ 2\pi } \sqrt { \frac { 1.{ \pi }^{ 2 } }{ 1 } } =\frac { \pi }{ 2\pi } =0.5\\$