JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Elasticity & SHM

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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 maximum energy in the spring are ______ Hz and ______ J, respectively. (Take

g = 10

m/s

^{2}

)

(A) $\frac{5 50}{π}$ and $8 \times 1 0^{- 3}$ ✓ Correct
(B) $\frac{5 50}{π}$ and 8
(C) $1050$ and $2 \times 1 0^{- 3}$
(D) $\frac{5 50}{π}$ and $16 \times 1 0^{- 3}$

Answer: (A)

Solution

The static stretch gives the stiffness:

k = \frac{m g}{x _{0}} = \frac{0.2 \times 10}{2 \times 1 0 ^{- 3}} = 1000 N/m

. The frequency is

f = \frac{1}{2 π} \frac{k}{m} = \frac{1}{2 π} \frac{1000}{0.2} = \frac{50 2}{2 π} = \frac{5 50}{π} Hz

. The mass oscillates with amplitude

2 mm

about the equilibrium stretch of

2 mm

, so the spring's greatest deformation is

4 mm

; the maximum energy stored in it is

\frac{1}{2} k x_{ma x}^{2} = \frac{1}{2} (1000) (4 \times 1 0^{- 3})^{2} = 8 \times 1 0^{- 3} J

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