JEE Main 2026 · 6 Apr · Shift 2 · Physics Q36

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PhysicsSingle correct+4 / -1Question 36 of 75

A spring stretches by 2 mm when it is loaded with a mass of 200 g. From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maxmimum energy in the spring are ________ Hz and ________ J, respectively.

(Take $g=10$ m/s²)

\dfrac{5\sqrt{50}}\pi

and

8\times10^{-3}

\dfrac{5\sqrt{50}}\pi

and 8

10\sqrt{50}

and

2\times10^{-3}

\dfrac{5\sqrt{50}}\pi

and

16\times10^{-3}

Answer and worked solution

Official answer: A

Spring constant. $k=\dfrac{0.2\times10}{0.002}=1000$ N/m.

Frequency.
$f=\frac1{2\pi}\sqrt{\frac{1000}{0.2}}=\frac{50\sqrt2}{2\pi}=\frac{5\sqrt{50}}\pi\text{ Hz}.$

Maximum spring energy. At the lowest point the extension is 4 mm:
$\tfrac12(1000)(0.004)^2=8\times10^{-3}\text{ J}.$

(A)

Topics:Spring-mass systems Energy in SHM

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JEE Main 2026 · 6 Apr · Shift 2, Physics question 36