Each of the two strings of length 51 . 6 cm and 49 . 1 cm are equally tensioned by 20 N force. Mass per unit length of both the strings is the same and equal to 1 g / m . When both the strings vibrate

Question

Each of the two strings of length 51 . 6 cm and 49 . 1 cm are equally tensioned by 20 N force. Mass per unit length of both the strings is the same and equal to 1 g / m . When both the strings vibrate simultaneously in fundamental mode the number of beats is--

TestHub · Accepted Answer

l 1 = 0.516 m , l 2 = 0. 491 m, T = 20 N . Mass per unit length, μ = 0.001 kg/m Frequency , v ⁡ = 1 2 l ⁡ T ⁡ μ v ⁡ 1 = 1 2 × 0 · 51 6 2 0 0 · 0 0 1 v ⁡ 2 = 1 2 × 0 · 4 9 1 2 0 0 · 0 0 1 ∴ Number of beats = υ 1 - υ 2 = 7

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