JEE Advanced 2022 Paper 2 · Q07 · Spring-Block
On a frictionless horizontal plane, a bob of mass $m = 0.1\,\text{kg}$ is attached to a spring with natural length $l_0 = 0.1\,\text{m}$. The spring constant is $k_1 = 0.009\,\text{N m}^{-1}$ when the length of the spring $l > l_0$ and is $k_2 = 0.016\,\text{N m}^{-1}$ when $l < l_0$. Initially the bob is released from $l = 0.15\,\text{m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T = (n\pi)\,\text{s}$, then the integer closest to $n$ is _____ .
Reveal answer + step-by-step solution
Correct answer:6
Solution
The bob executes simple harmonic motion with two different stiffness regimes. The half-period in the regime $l > l_0$ uses $k_1$: $T_1/2 = \pi\sqrt{m/k_1}$. The half-period in $l < l_0$ uses $k_2$: $T_2/2 = \pi\sqrt{m/k_2}$. The full oscillation period $T = \pi(\sqrt{m/k_1} + \sqrt{m/k_2})$. Substituting $m = 0.1$, $k_1 = 0.009$, $k_2 = 0.016$: $\sqrt{0.1/0.009} = \sqrt{100/9} = 10/3$, and $\sqrt{0.1/0.016} = \sqrt{100/16} = 10/4 = 2.5$. So $T = \pi(10/3 + 5/2) = \pi(20/6 + 15/6) = (35/6)\pi \approx 5.83\pi$. The integer closest to $n = 35/6$ is $6$.
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →