Question

Problem 2 [8 pts] Oscillator As a quality control technician at a violin string factory, you cut a sample of E-string off a large roll. The sample that you cut has a mass of ms = 1.021 grams and a full length of 2.5 meters. To test the string, you stretch some of it across a length L = 0.35 m, applying tension by means of a hanging mass m (as pictured). A variable frequency oscillator is used to excite standing waves. (a) What mass must you hang to have the fundamental standing wave vibrate at middle C (261.626 Hz)? (b) What is the speed of the wave on the string? (c) You find that the fundamental standing wave is excited at 262.6 Hz. Assuming the oscillator is high quality (accurate), which of l, m, and m, is the most likely culprit for an error? Hint: write a symbolic expression for the frequency and consider propagation of errors. (d) Sketch the fundamental wave as well as the next three harmonics (2nd through 4th). Tabulate the wavelengths and frequencies. (e) For the 4th harmonic, the full height from top to bottom of the antinode of the standing wave is 5.0 mm. Write an expression for one of the traveling waves that results in this standing wave. Provide the amplitude, speed, wavelength, and frequency

Answer 1

To have the fundamental standing wave: a) A2L 2(0.35) 0.70 m L= 2 The frequency: f 261.626 Hz = The speed of the wave is: vAf 0.7 261.626 183.1382 The speed is also giving by: FT where Fr mg and u ms v = mgl v = ms Solving for m m,v2 gls The speed is: (1.021 x 10-3)(183.1382)2 = 1.40 kg m= 9.8 2.5 b) т v=183 S The frequency is giving by: c) 1 mgl and A 2L f - = Then: mgls 1 f = 2L т The uncertainty of the frequency is: 1 Δm 1 ΔΙ 1 Am [AL Af f 2 L 2 m 2 m Then, the higher contribution is L d) For each mode: 2L= 0.35 m 2L = 0.233 m = 3 0.175 m 2 = 2 261.626 = 523.252 Hz f2 2fi fs 3f 3 261.626 = 784.878 Hz f4 4f 4* 261.626 = 1046.504 Hz