# Three pulley system

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

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

Note by Tanishq Varshney
3 years, 7 months ago

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- 3 years, 7 months ago

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

- 3 years, 7 months ago

Yeah.

- 3 years, 7 months ago

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

- 3 years, 7 months ago

- 3 years, 7 months ago

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

- 3 years, 7 months ago

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

- 2 years, 1 month ago