JEE Advanced 2022 — Paper 2 — PHY-I solution

JEE Advanced2022Paper 2PHY-I

On a frictionless horizontal plane, a bob of mass $m = 0.1$ kg is attached to a spring with natural length $l_{0} = 0.1$ m. The spring constant is $k_{1} = 0.009$ Nm $^{- 1}$ when the length of the spring $l > l_{0}$ and is $k_{2} = 0.016$ Nm $^{- 1}$ when $l < l_{0}$ . Initially the bob is released from $l = 0.15$ m. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T = (nπ)$ s, then the integer closest to $n$ is _______.

Your answer

Solution

Each half-oscillation about the natural length is half the period of an SHM with the appropriate spring constant.

Half-period with $k_{1}$ (extended side): $π m / k_{1} = π 0.1/0.009 = \frac{π}{0.3}$ s.

Half-period with $k_{2}$ (compressed side): $π m / k_{2} = π 0.1/0.016 = \frac{π}{0.4}$ s.

Total period $T = \frac{π}{0.3} + \frac{π}{0.4} = π (\frac{4 + 3}{1.2}) = \frac{7 π}{1.2} = \frac{70 π}{12} \approx 5.83 π$ s.

The integer closest to $n = 5.83$ is $6$ .

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