A mass 0.10 kg hangs from an elastic rod of length 2 m, cross-sectional area 1 mm² and Young's modulus 2×10¹¹ Pa. The rod mass is negligible. If the angular frequency of small longitudinal oscillations is N rad/s, find N.
Physics · Simple Harmonic Motion · time period of linear shm
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Question

A mass 0.10 kg hangs from an elastic rod of length 2 m, cross-sectional area 1 mm² and Young's modulus 2×10¹¹ Pa. The rod mass is negligible. If the angular frequency of small longitudinal oscillations is N rad/s, find N.

Physics · Simple Harmonic Motion · time period of linear shm

G1 Physics · Accepted Answer

The elastic rod has spring constant k = YA/L. Substitute Y = 2×10¹¹ Pa, A = 1 mm² = 10⁻⁶ m², L = 2 m. Then, k = (2×10¹¹)(10⁻⁶)/2 = 10⁵ N/m. Therefore, ω = √(k/M) = √(10⁵/0.10) = √10⁶ = 1000 rad/s. Hence, N = 1000.