Topic: Theory of Machines
The reading of a spring balance is from 0 to 200 N and is 10 cm long. A body suspended from the spring balance is observed to oscillate vertically at 2 Hz. The mass of the body is nearly
- A. 22.5 kg
- B. 12.5 kg
- C. 37 kg
- D. 45 kg
Answer: B. 12.5 kg
Explanation: The stiffness (k) of the spring is calculated as Force divided by deflection: k = 200 N / 0.1 m = 2000 N/m. The natural frequency of the oscillation is given by f = (1 / (2 * pi)) * $\sqrt{k / m}$. Substituting the known frequency f = 2 Hz, we get 2 = (1 / (2 * pi)) * $\sqrt{2000 / m}$, which simplifies to 4 * pi = $\sqrt{2000 / m}$. Squaring both sides yields 16 * pi² = 2000 / m, resulting in m = 2000 / (16 * pi²). This evaluates to approximately 12.66 kg, which closely matches 12.5 kg.