Question Detail
Question
A string is damped 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
For a string fixed at both ends, a standing wave with 'n' loops has a length L given by L = nλ/2, where λ is the wavelength.Given: n = 5 loops.Velocity of wave on a string (v) is given by v = √(T/μ), where T is tension and μ is mass per unit length.Given: T = 2.5 N, μ = 0.001 kg/m (assuming N/m is a typo for kg/m).v = √(2.5 N / 0.001 kg/m) = √(2500) m/s = 50 m/s.The relationship between wave velocity (v), frequency (f), and wavelength (λ) is v = fλ.Given: f = 750 Hz.λ = v/f = 50 m/s / 750 Hz = 1/15 m.Now, calculate the length of the string (L):L = nλ/2 = 5 * (1/15 m) / 2 = 5/30 m = 1/6 m.L ≈ 0.1666... m, which rounds to 0.167 m.
Hint
First, calculate the wave velocity using tension and mass per unit length. Then find the wavelength using frequency and velocity. Finally, relate the wavelength to the length of the string for a standing wave with 'n' loops.