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?

## Comments

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

\( k \) is constant, as the restoring force is assumed to not change. – Ameya Daigavane · 6 months ago

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question? – Anandhu Raj · 6 months ago

Thank you very much, Sir! Do you have a problem if I mention your name in thisLog in to reply

– Ameya Daigavane · 6 months ago

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

– Ameya Daigavane · 6 months ago

Absolutely not, thank you for the honour!Log in to reply