When a mass m is hung on a spring, the spring stretched by 6 cm. If the loaded spring is pulled downward a little and released, then the period of vibration of the system will be

When a mass m is hung on a spring, it reaches equilibrium where the gravitational force (mg) equals the spring's restoring force (kΔx) [t4][t5]. According to Hooke's Law, mg = kΔx, which allows us to express the ratio m/k as Δx/g [t6]. The period of vibration (T) for a mass-spring system undergoing simple harmonic motion is given by the formula T = 2π√(m/k) [t2]. Substituting the equilibrium condition, the formula becomes T = 2π√(Δx/g). Given the extension Δx = 6 cm = 0.06 m and using the standard acceleration due to gravity g ≈ 9.8 m/s², the calculation is T = 2π√(0.06 / 9.8). This yields T ≈ 2π√(0.006122) ≈ 2 × 3.14159 × 0.0782 ≈ 0.491 seconds. Therefore, the period of vibration is approximately 0.49 s, matching option 3.