Question Detail
Question
A string is clamped at two ends and set into vibration such that the standing wave has five loops. A tuning fork of frequency 750 Hz sets the string into vibration. It is observed that the tension in the string is 2.5 N and the mass per unit length 0.001 N/m. If the amplitude of the wave is 1.8 mm, find the length of the string.
Correct Answer
Option B is the correct answer.
Detailed Explanation
Five loops means the 5th harmonic (n=5). The frequency is f = (n/2L)√(T/μ). μ = 0.001 kg/m (since 0.001 N/m is the weight per unit length, mass per unit length is μ = weight per unit length / g = 0.001/9.8 ≈ 0.000102 kg/m. But the question gives mass per unit length as 0.001 N/m, which is weight per unit length. So μ = 0.001/g = 0.001/9.8 = 1.02e-4 kg/m. Using f = 750, n=5, T=2.5, μ=1.02e-4. 750 = (5/(2L))√(2.5/1.02e-4) = (5/(2L))√(24509.8) = (5/(2L))×156.56 = (391.4)/L. L = 391.4/750 = 0.522 m. None of the options match. If μ = 0.001 kg/m (mass per unit length), then 750 = (5/(2L))√(2.5/0.001) = (5/(2L))√2500 = (5/(2L))×50 = 125/L => L = 125/750 = 0.167 m.
Hint
Use the formula for the frequency of a standing wave on a string: f = (n/2L)√(T/μ).