Practice question

Given: Two identical springs ( k = 75 N/m ) are attached to a 1.5 kg mass as in Fig. 13.14. What is the frequency? These values define the system as per NCERT data. Formula: Effective k_{eff = 2k = 2 × 75 = 150 N/m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: omega = sqrtfrack_{effm = sqrt150/1.5 = sqrt100 = 10 rad/s . v = omega/2π = 10/2 × 3.14 approx 1.59 Hz . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.