The pulley, strings and springs are light.I just wanted to confirm my answer (although I know I am wrong) that

\[\Large{\omega^2=\frac{4}{M \left(\frac{1}{k_{1}}+\frac{2}{k_{2}}+\frac{4}{k_{3}}\right)}}\]

The pulley, strings and springs are light.I just wanted to confirm my answer (although I know I am wrong) that

\[\Large{\omega^2=\frac{4}{M \left(\frac{1}{k_{1}}+\frac{2}{k_{2}}+\frac{4}{k_{3}}\right)}}\]

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TopNewesti'm getting \(\large\frac{4}{M(\frac{1}{k_{1}}+\frac{1}{4k_{2}}+\frac{1}{16k_3})}\) – Keshav Tiwari · 1 year ago

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– Pulkit Gupta · 1 year ago

Yeah.Log in to reply

@Abhishek Singh @Abhineet Nayyar @Abhishek Sharma @Aditya Kumar – Tanishq Varshney · 1 year ago

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– Tanishq Varshney · 1 year ago

Sorry I don't have its answer can u post your method/solution please!Log in to reply

@Keshav Tiwari @Aditya Kumar @Abhineet Nayyar plz post your method. – Tanishq Varshney · 1 year ago

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Put \({ x }_{ 1 }=4a/{ k }_{ 1 }\), \({ x }_{ 2 }=2a/{ k }_{ 2 }\), and \({ x }_{ 3 }=a/{ k }_{ 3 }\). Also, put \(x=a/k\) where k is the net inertia factor and \(x, k\) will also follow the same relation as the other pairs.

Solving this: \({ \omega }^{ 2 }=\frac { 8 }{ M(\frac { 16 }{ { k }_{ 1 } } +\frac { 4 }{ { k }_{ 2 } } +\frac { 1 }{ { k }_{ 3 } } ) } \) – Abhineet Nayyar · 1 year ago

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I'm getting \(\frac { 8 }{ M(\frac { 16 }{ { k }_{ 1 } } +\frac { 4 }{ { k }_{ 2 } } +\frac { 1 }{ { k }_{ 3 } } ) } \) – Abhineet Nayyar · 1 year ago

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