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

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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

Log in to reply

Yeah.

Log in to reply

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

Log in to reply

Comment deleted Jan 08, 2016

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply