Q61 / 75 · JEE Main 2026 (06 April Shift 2) · Physics
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$^2$)
- Correct answerA$\dfrac{5\sqrt{50}}{\pi}$ and $8 \times 10^{-3}$
- B$\dfrac{5\sqrt{50}}{\pi}$ and $8$
- C$10\sqrt{50}$ and $2 \times 10^{-3}$
- D$\dfrac{5\sqrt{50}}{\pi}$ and $16 \times 10^{-3}$
Solution
At equilibrium, the restoring force of the spring balances the weight of the mass:
$mg = kx_0$
$k = \dfrac{mg}{x_0} = \dfrac{0.2 \times 10}{2 \times 10^{-3}} = 1000$ N/m
The frequency of oscillation is given by:
$f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$
$f = \dfrac{1}{2\pi} \sqrt{\dfrac{1000}{0.2}} = \dfrac{1}{2\pi} \sqrt{5000} = \dfrac{10\sqrt{50}}{2\pi} = \dfrac{5\sqrt{50}}{\pi}$ Hz
The maximum energy in the spring corresponds to its maximum elastic potential energy, which occurs at the maximum extension.
Maximum extension $x_{max} = x_0 + A = 2 \text{ mm} + 2 \text{ mm} = 4 \text{ mm} = 4 \times 10^{-3}$ m
Maximum energy in the spring $U_{max} = \dfrac{1}{2} k x_{max}^2$
$U_{max} = \dfrac{1}{2} \times 1000 \times (4 \times 10^{-3})^2$
$U_{max} = 500 \times 16 \times 10^{-6} = 8000 \times 10^{-6} = 8 \times 10^{-3}$ J
Answer: $\dfrac{5\sqrt{50}}{\pi}$ and $8 \times 10^{-3}$
$mg = kx_0$
$k = \dfrac{mg}{x_0} = \dfrac{0.2 \times 10}{2 \times 10^{-3}} = 1000$ N/m
The frequency of oscillation is given by:
$f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$
$f = \dfrac{1}{2\pi} \sqrt{\dfrac{1000}{0.2}} = \dfrac{1}{2\pi} \sqrt{5000} = \dfrac{10\sqrt{50}}{2\pi} = \dfrac{5\sqrt{50}}{\pi}$ Hz
The maximum energy in the spring corresponds to its maximum elastic potential energy, which occurs at the maximum extension.
Maximum extension $x_{max} = x_0 + A = 2 \text{ mm} + 2 \text{ mm} = 4 \text{ mm} = 4 \times 10^{-3}$ m
Maximum energy in the spring $U_{max} = \dfrac{1}{2} k x_{max}^2$
$U_{max} = \dfrac{1}{2} \times 1000 \times (4 \times 10^{-3})^2$
$U_{max} = 500 \times 16 \times 10^{-6} = 8000 \times 10^{-6} = 8 \times 10^{-3}$ J
Answer: $\dfrac{5\sqrt{50}}{\pi}$ and $8 \times 10^{-3}$
Hence, the correct answer is option: 'A'.