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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

The force constant of two springs are

\;{K_1}

and

{K_2}

Both are stretched till their elastic energies are equal. If the stretching forces are

{F_1}

and

{F_2},

then

{F_1}:{F_2}

is

View answer and explanation

Correct option: C

\sqrt {{K_1}} :\sqrt {{K_2}}

Option C (

\sqrt {{K_1}} :\sqrt {{K_2}}

) is correct because it is derived from equating the elastic potential energies and relating them to the stretching forces.

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