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A uniform spring whose initial length is $l$ has force const $K$ . Spring is cut into 2 pieces of length $l_{1}$ & $l_{2}$ such that $l_{1} = nl_{2}$ , $n$ is integer. Now mas ' $m$ ' is attached to first spring. The time period of its oscillation will be

Option 1)
$T = 2\pi \sqrt{\frac{mn}{K(n+1)}}$

Option 2)
$T = 2\pi \sqrt{\frac{m}{K(n+1)}}$

Option 3)
$T = 2\pi \sqrt{\frac{m}{nK}}$

Option 4)
$T = 2\pi \sqrt{\frac{m(n+1)}{nK}}$

Answers (1)

best_answer

$K\alpha \frac{1}{l} \Rightarrow K' = \left ( \frac{n+1}{n} \right )K$

$\Rightarrow T = 2\pi\sqrt{\frac{mn}{K(n+1)}}$

Cutting of Spring -

If a spring is cut into 2 peices of length $l_{1}$ & $l_{2}$ where $l_{1} = nl_{2}$

Spring const of

* First part $K_{1} = \frac{K(n+1)}{n}$

$K_{2} = (n+1)K$

- wherein

Formula

$\frac{K_{1}}{K_{2}} = \frac{1}{n}$

Option 1)

$T = 2\pi \sqrt{\frac{mn}{K(n+1)}}$

This is correct.

Option 2)

$T = 2\pi \sqrt{\frac{m}{K(n+1)}}$

This is incorrect.

Option 3)

$T = 2\pi \sqrt{\frac{m}{nK}}$

This is incorrect.

Option 4)

$T = 2\pi \sqrt{\frac{m(n+1)}{nK}}$

This is incorrect.

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prateek

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