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, 10 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Log in to reply

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, 10 months ago

Log in to reply

Yeah.

Pulkit Gupta - 3 years, 10 months ago

Log in to reply

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, 10 months ago

Log in to reply

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

Tanishq Varshney - 3 years, 10 months ago

Log in to reply

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, 10 months ago

Log in to reply

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

Gabriele Manganelli - 2 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...