Two small bodies of mass 2 kg each hang on a spring of force constant of 200 N / m , attached to each other using a thread with a length of 10 cm, as shown in the figure. We burn the thread. Find the

Question

Two small bodies of mass 2 kg each hang on a spring of force constant of $200 N / m$ , attached to each other using a thread with a length of 10 cm, as shown in the figure. We burn the thread. Find the distance between the two bodies (in cm) when the top body first arrives at its highest position. If the distance comes out to be $16 \times n$ , fill n [Take: $g = 10 m / s^{2}$ and $π^{2} = 10$ ]

TestHub · Accepted Answer

After the thread burns, the top body does SHM on the spring: $T = 2 π \frac{m}{k} = 2 π \frac{2}{200} = \frac{π}{5} s$

The equilibrium shifts up by mg/k, so the amplitude is $A = \frac{2 g}{200} = 0.1 m$ .

The top body first reaches its highest point after T/2, having risen 2A = 0.2 m.

Meanwhile the bottom body falls freely for T/2 = π/10 s: $h = \frac{1}{2} (10) (\frac{π}{10})^{2} = \frac{π ^{2}}{20} = 0.5 m$

Total separation = 2A + thread length + h = 0.2 + 0.1 + 0.5 = 0.8 m = 80 cm = 16 x 5, so n = 5.

Answer: 5

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