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# 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
1 year, 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.

- 1 year, 8 months ago

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

- 1 year, 8 months ago

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

- 1 year, 8 months ago

Absolutely not, thank you for the honour!

- 1 year, 8 months ago