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#stiffness

2 public questions tagged with this topic.

Which factor in a spring-mass system directly influences the stiffness of the restoring force?

**Spring-mass system** has period T = 2π√(m/k), frequency f = (1/2π)√(k/m), ω = √(k/m), where k spring constant (N/m) and m mass (kg). For parallel combination, k_eff = k₁ + k₂, series gives 1/k_eff = 1/k₁ + 1/k₂, affecting ω = √(k_eff/m) and T = 2π√(m/k_eff). The spring constant ( k in F = -kx ) determines the stiffness, as it measures the force per unit displacement, affecting the system’s oscillation characteristics. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Spring constant follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

Why do engineers prefer materials with a large Young’s modulus for designing columns and beams?

Materials with a large Young’s modulus are stiffer, requiring greater force to produce small deformations, which ensures better resistance to bending and stretching. As per NCERT, applying relevant law/formula with correct units and sign convention leads to They provide greater stiffness. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.