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
For a string clamped at both ends, a standing wave with 'n' loops corresponds to the 'n'th harmonic. The length of the string (L) for 'n' loops is given by L = n(λ/2), where λ is the wavelength.Here, n = 5 loops, so L = 5(λ/2) => λ = 2L/5.The frequency of vibration (f) = 750 Hz.The tension (T) = 2.5 N.The mass per unit length (μ) = 0.001 N/m. (Assuming N/m is a typo and should be kg/m, as mass per unit length is typically in kg/m). So, μ = 0.001 kg/m.The speed of a wave on a string (v) is given by v = √(T/μ).v = √(2.5 N / 0.001 kg/m) = √(2500) m/s = 50 m/s.Also, the wave speed is related to frequency and wavelength by v = fλ.Substituting λ = 2L/5 into v = fλ:v = f(2L/5).Rearranging for L: L = (5v) / (2f).L = (5 × 50 m/s) / (2 × 750 Hz) = 250 / 1500 m = 1/6 m.L ≈ 0.16666... m, which rounds to 0.167 m.
Hint
First, calculate the wave speed on the string using tension and linear mass density. Then, use the relationship between wave speed, frequency, and wavelength, along with the condition for standing waves (number of loops and wavelength).