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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## Comments

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TopNewesti'm getting \(\large\frac{4}{M(\frac{1}{k_{1}}+\frac{1}{4k_{2}}+\frac{1}{16k_3})}\)

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Yeah.

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@Abhishek Singh @Abhineet Nayyar @Abhishek Sharma @Aditya Kumar

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Comment deleted Jan 08, 2016

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

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@Keshav Tiwari @Aditya Kumar @Abhineet Nayyar plz post your method.

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Take the tension with the string attached to the mass as \(8T\). Now, the following strings will have half the tension as the previous one, as you already know. Consider the mass moves by a distance \(x\). The spring \({ k }_{ 1 }\) gets stretched by \({ x }_{ 1 }\), \({ k }_{ 2 }\) with \({ x }_{ 2 }\), and \({ k }_{ 3 }\) with \({ x }_{ 3 }\). You can clearly see, that \({ k }_{ 1 }\times { x }_{ 1 }=4T\), \({ k }_{ 2 }\times { x }_{ 2 }=2T\) and \({ k }_{ 3 }\times { x }_{ 3 }=T\) Also by virtual work done: \(8x=4{ x }_{ 1 }+2{ x }_{ 2 }+{ x }_{ 3 }\)

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: \({ \omega }^{ 2 }=\frac { 8 }{ M(\frac { 16 }{ { k }_{ 1 } } +\frac { 4 }{ { k }_{ 2 } } +\frac { 1 }{ { k }_{ 3 } } ) } \)

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Are you sure that you didn't exchange the indexes 1 and 3?

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I'm getting \(\frac { 8 }{ M(\frac { 16 }{ { k }_{ 1 } } +\frac { 4 }{ { k }_{ 2 } } +\frac { 1 }{ { k }_{ 3 } } ) } \)

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