# How to do such questions?

I have doubt regarding the solution of the following question,

"Block $A$ of mass $m$ is performing SHM of amplitude $a$. Another block $B$ of mass $m$ is gently placed on A when it passes through mean position and $B$ sticks to $A$. Find amplitude of new SHM."

Given solution,

For mass $m:\frac { 1 }{ 2 } m{ u }^{ 2 }=\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 }$ ----$\boxed{1}$

For mass $2m:\frac { 1 }{ 2 } 2m{ v }^{ 2 }=\frac { 1 }{ 2 } 2m\left( \frac { \omega }{ \sqrt { 2 } } \right) ^{ 2 }{ A }^{ 2 }$

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

$\therefore \frac { 1 }{ 2 } 2m{ \left( \frac { u }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 2 } 2m\left( \frac { \omega }{ \sqrt { 2 } } \right) ^{ 2 }{ A }^{ 2 }$ ----$\boxed{2}$

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

$4=\frac { 2{ a }^{ 2 } }{ { A }^{ 2 } }$

New amplitude, $A=\frac { a }{ \sqrt { 2 } }$

What I don't get is that how $\frac { \omega }{ \sqrt { 2 } }$ came in step $\boxed{2}$. Could someone please help? Note by Anandhu Raj
4 years, 8 months ago

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$\omega = \sqrt{\dfrac{k}{m}}$ where $F = - kx = ma$ is the restoring force.

$k$ is constant, as the restoring force is assumed to not change.

- 4 years, 8 months ago

Thank you very much, Sir! Do you have a problem if I mention your name in this question?

- 4 years, 8 months ago

Absolutely not, thank you for the honour!

- 4 years, 8 months ago

Also, I think you may have accidentally posted the same note many times, just delete the other ones.

- 4 years, 8 months ago