A tribute of Ameya Daigavane

Classical Mechanics Level pending

Block A A of mass m m is performing SHM of amplitude a a . Another block B B of mass m m is gently placed on A when it is at a 2 \frac { a }{ 2 } from mean position and B B sticks to A A . Find amplitude of new SHM.

Details Given a = 1 a=1

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The answer is 0.79.

Anandhu Raj by Anandhu Raj , 23, India

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1 solution

Anandhu Raj
Mar 26, 2016

Kinetic energies at a 2 \frac { a }{ 2 }

For mass m : 1 2 m u 2 = 1 2 m ω 2 [ a 2 − ( a 2 ) 2 ] m:\frac { 1 }{ 2 } m{ u }^{ 2 }=\frac { 1 }{ 2 } m{ \omega }^{ 2 }\left[ { a }^{ 2 }-{ \left( \frac { a }{ 2 } \right) }^{ 2 } \right] ---- 1 \boxed{1} For mass 2 m : 1 2 2 m v 2 = 1 2 2 m ( ω 2 ) 2 [ A 2 − ( a 2 ) 2 ] 2m:\frac { 1 }{ 2 } 2m{ v }^{ 2 }=\frac { 1 }{ 2 } 2m\left( \frac { \omega }{ \sqrt { 2 } } \right) ^{ 2 }\left[ { A }^{ 2 }-{ \left( \frac { a }{ 2 } \right) }^{ 2 } \right]

By conservation of momentum, v = u 2 v=\frac { u }{ 2 }

∴ 1 2 2 m ( u 2 ) 2 = 1 2 2 m ( ω 2 ) 2 [ A 2 − ( a 2 ) 2 ] \therefore\frac { 1 }{ 2 } 2m{ \left( \frac { u }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 2 } 2m\left( \frac { \omega }{ \sqrt { 2 } } \right) ^{ 2 }\left[ { A }^{ 2 }-{ \left( \frac { a }{ 2 } \right) }^{ 2 } \right] ---- 2 \boxed{2}

Dividing 1 \boxed{1} by 2 \boxed{2}

2 = a 2 − ( a 2 ) 2 A 2 − ( a 2 ) 2 2=\frac { { a }^{ 2 }-{ \left( \frac { a }{ 2 } \right) }^{ 2 } }{ { A }^{ 2 }-{ \left( \frac { a }{ 2 } \right) }^{ 2 } }

Solving gives A = a 5 8 A=a\sqrt { \frac { 5 }{ 8 } }

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