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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

A uniform spring of normal length

\lambda

has a force constant

\mathrm{k}

. It is cut into two pieces of lengths

\lambda_1

and

\lambda_2

such that

\lambda_1=

\mathrm{n} \lambda_2

, where

\mathrm{n}

is an integer. Then the value of

\mathrm{k}_1

(force constant of spring of length

\lambda_1

) is -

View answer and explanation

Correct option: B

\frac{\mathrm{k}(\mathrm{n}+1)}{\mathrm{n}}

Option B (

\frac{\mathrm{k}(\mathrm{n}+1)}{\mathrm{n}}

) is correct because the force constant of a spring is inversely proportional to its length, and the total length is divided proportionally.

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