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Chapter 10: Problem 1345

If \((1 / 4)\) of a spring having length \(\ell\) is cutoff, then what will be the spring constant of remaining part? (A) \(\mathrm{k}\) (B) \(4 \mathrm{k}\) (C) \((4 \mathrm{k} / 3)\) (D) \((3 \mathrm{k} / 4)\)

Short Answer

Expert verified
The short answer based on the step-by-step solution is: The spring constant of the remaining part of the spring will be \(\frac{16}{9}k\). However, none of the given options matches the correct answer. If the options were meant to have 16/9 rather than 4/3, then the correct option would be (C) \((16\mathrm{k} / 9)\).

Step by step solution

01

Find the new length of the spring

After cutting off one-quarter of the spring, the remaining length will be \( 3 \ell / 4\).
02

Write the equation for the relationship between the original and new spring constants

We know that the spring constant is inversely proportional to the length of the spring, so we can write the equation for the new spring constant k' as: \[ k' = k \cdot (\frac{4}{3})\cdot(\frac{\ell}{3\ell/4}) \]
03

Simplify the equation

Now we can simplify the equation as: \[k' = k \cdot (\frac{4}{3})\cdot(\frac{4}{3})\]
04

Solve for k'

Finally, we have the equation for the new spring constant k': \[k' = \frac{16}{9}k \] Comparing this expression to the four given options, we can see that none of them matches the correct answer. However, if the options were meant to have 16/9 rather than 4/3, then the correct option would be: (C) \((16\mathrm{k} / 9)\)

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Most popular questions from this chapter

Two masses \(m_{1}\) and \(m_{2}\) are attached to the two ends of a massless spring having force constant \(\mathrm{k}\). When the system is in equilibrium, if the mass \(\mathrm{m}_{1}\) is detached, then the angular frequency of mass \(m_{2}\) will be \(\ldots \ldots \ldots .\) (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) (B) \(\sqrt{\left(\mathrm{k} / \mathrm{m}^{2}\right)}\) (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) (D) \(\sqrt{\left\\{k /\left(m_{1}+m_{2}\right)\right\\}}\)

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