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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 k1 gets stretched by x1, k2 with x2, and k3 with x3.
You can clearly see, that k1×x1=4T, k2×x2=2T and k3×x3=T
Also by virtual work done:
8x=4x1+2x2+x3
Put x1=4a/k1, x2=2a/k2, and x3=a/k3.
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.
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:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
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\(
...\)
or\[
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
@Abhishek Singh @Abhineet Nayyar @Abhishek Sharma @Aditya Kumar
i'm getting M(k11+4k21+16k31)4
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Yeah.
I'm getting M(k116+k24+k31)8
@Keshav Tiwari @Aditya Kumar @Abhineet Nayyar plz post your method.
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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 k1 gets stretched by x1, k2 with x2, and k3 with x3. You can clearly see, that k1×x1=4T, k2×x2=2T and k3×x3=T Also by virtual work done: 8x=4x1+2x2+x3
Put x1=4a/k1, x2=2a/k2, and x3=a/k3. 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: ω2=M(k116+k24+k31)8
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Are you sure that you didn't exchange the indexes 1 and 3?