Three pulley system

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

ω2=4M(1k1+2k2+4k3)\Large{\omega^2=\frac{4}{M \left(\frac{1}{k_{1}}+\frac{2}{k_{2}}+\frac{4}{k_{3}}\right)}}

Note by Tanishq Varshney
3 years, 7 months ago

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i'm getting 4M(1k1+14k2+116k3)\large\frac{4}{M(\frac{1}{k_{1}}+\frac{1}{4k_{2}}+\frac{1}{16k_3})}

Keshav Tiwari - 3 years, 7 months ago

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Yeah.

Pulkit Gupta - 3 years, 7 months ago

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I'm getting 8M(16k1+4k2+1k3)\frac { 8 }{ M(\frac { 16 }{ { k }_{ 1 } } +\frac { 4 }{ { k }_{ 2 } } +\frac { 1 }{ { k }_{ 3 } } ) }

Abhineet Nayyar - 3 years, 7 months ago

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

Tanishq Varshney - 3 years, 7 months ago

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Take the tension with the string attached to the mass as 8T8T. Now, the following strings will have half the tension as the previous one, as you already know. Consider the mass moves by a distance xx. The spring k1{ k }_{ 1 } gets stretched by x1{ x }_{ 1 }, k2{ k }_{ 2 } with x2{ x }_{ 2 }, and k3{ k }_{ 3 } with x3{ x }_{ 3 }. You can clearly see, that k1×x1=4T{ k }_{ 1 }\times { x }_{ 1 }=4T, k2×x2=2T{ k }_{ 2 }\times { x }_{ 2 }=2T and k3×x3=T{ k }_{ 3 }\times { x }_{ 3 }=T Also by virtual work done: 8x=4x1+2x2+x38x=4{ x }_{ 1 }+2{ x }_{ 2 }+{ x }_{ 3 }

Put x1=4a/k1{ x }_{ 1 }=4a/{ k }_{ 1 }, x2=2a/k2{ x }_{ 2 }=2a/{ k }_{ 2 }, and x3=a/k3{ x }_{ 3 }=a/{ k }_{ 3 }. Also, put x=a/kx=a/k where k is the net inertia factor and x,kx, k will also follow the same relation as the other pairs.

Solving this: ω2=8M(16k1+4k2+1k3){ \omega }^{ 2 }=\frac { 8 }{ M(\frac { 16 }{ { k }_{ 1 } } +\frac { 4 }{ { k }_{ 2 } } +\frac { 1 }{ { k }_{ 3 } } ) }

Abhineet Nayyar - 3 years, 7 months ago

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

Gabriele Manganelli - 2 years, 1 month ago

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