A 0.500 g wire is stretched between two points 95.0 cm apart. If the tension in the wire is 600 N, find the wire's first, second, and third harmonics.______Hz (1st)______Hz (2nd)______Hz (3rd)
Kia ora In order to answer this question, we first need to ascertain the speed of a wave in this wire. The speed of a wave on a wire depends upon the tension 'T' and the linear density (mass per unit length) 'μ' of the wire. v=√(T/μ) T=600 N The linear density of your wire is 5.00E-4 kg/0.95=5.263E-4 kg/m. So for this wire, v=√(600/5.263E-4) =1068 m/s Now we have the speed, we need to find the wavelengths that correspond to the harmonics we are interested in. The wire is fixed at both ends. Nodes occur at fixed ends. Therefore the first harmonic will occur when there is a node at each end and the longest possible wavelength that satisfies this condition is λ=2L (you get half a wavelength on the wire). So λ=2*0.95=λ=1.90m. If v=1068m/s and λ=1.90 then by the wave equation v=fλ f=v/λ=1068/1.90=562.1 Hz. So that is the first harmonic. The second harmonic has twice the frequency of the first; the third harmonic has three times the frequency of the first. The second harmonic will therefore occur at 562.1*2=1124 Hz and the third at 562.1*3=1686 Hz. Because your data was given to 3sf you need to round your answer to 3sf: 1st harmonic: 562 Hz 2nd harmonic: 1120 Hz 3rd harmonic: 1680 Hz
