Question

A spring, of spring constant k, has two masses m₁ and m₂ , attached at its lower end. When the system is in equilibrium, the mass m₁ is gently removed. The amplitude, and time period, of the resulting oscillations of m₂ , would be, respectively

(1) \((\frac{m_1g}{k})\) and 2π \(\sqrt{\frac{m_1}{k}}\)

(2) \((\frac{m_2g}{k})\) and 2π \(\sqrt{\frac{m_2}{k}}\)

(3) \((\frac{m_1g}{k})\) and 2π \(\sqrt{\frac{m_2}{k}}\)

(4) \((\frac{m_2g}{k})\) and 2π \(\sqrt{\frac{m_1}{k}}\)

Answer 1

 (3) \((\frac{m_1g}{k})\) and 2π \(\sqrt{\frac{m_2}{k}}\)

The equilibrium extension, say y, of the spring, when both the masses m₁ and m₂ are suspended from it, equals (m₁ + m₂) g/k.

On removing the mass m1 , the equilibrium extension would be only (m₂g/k). Hence when the mass m¹ is gently removed, the mass m₂ would have an unbalanced extension (m₁g/k). This would, therefore, be the ampitude of its oscillatinos. 

Hence A = (m₁g/k)

The restoring force, on mass m₂ is equal to ky₁ (= M₁g) where y₁ would have been the extension of the spring under the mass m₁ alone. Hence its equation of motion is

∴ The oscillatinos of m₂ are simple harmonic with ω² = k/m₂.