# 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
2 years, 6 months ago

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

- 2 years, 6 months ago

Yeah.

- 2 years, 6 months ago

- 2 years, 6 months ago

Comment deleted Jan 08, 2016

- 2 years, 6 months ago

- 2 years, 6 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: $${ \omega }^{ 2 }=\frac { 8 }{ M(\frac { 16 }{ { k }_{ 1 } } +\frac { 4 }{ { k }_{ 2 } } +\frac { 1 }{ { k }_{ 3 } } ) }$$

- 2 years, 6 months ago

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

- 1 year ago

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

- 2 years, 6 months ago