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 kg/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 has a length L given by L = n * (λ/2), where λ is the wavelength. Here, n=5, so L = 5 * (λ/2).First, calculate the wave speed (v) using the tension (T) and mass per unit length (μ): v = √(T/μ) = √(2.5 N / 0.001 kg/m) = √(2500) = 50 m/s.Next, calculate the wavelength (λ) using the wave speed and frequency (f): λ = v/f = 50 m/s / 750 Hz = 1/15 m.Finally, calculate the length of the string: L = 5 * ( (1/15 m) / 2 ) = 5 * (1/30 m) = 5/30 m = 1/6 m ≈ 0.1667 m. The closest option is 0.167 m.Note: The unit 'N/m' for mass per unit length in the original question is a typo; it should be 'kg/m'.
Hint
First find the wave speed using tension and mass per unit length, then find the wavelength using frequency, and finally determine the string length for a standing wave with five loops.