Question

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. Find a) the fundamental frequency, and b) the speed of the waves on the string.

Answer 1

`\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}`

Step-by-step explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

`\frac{\text{23.40 Hz}}{\text{15.60 Hz}} = (1.500)/(1) \approx (3)/(2)`

`f = \frac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}`

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

`\begin{array}{rcl}L & = & (\lambda)/(2)\\\\\text{1.30 m} & = & (\lambda)/(2)\\\\\lambda & = & \text{2.60 m}\\\end{array}`

(b) Calculate the speed s

`\begin{array}{rcl}v_(1)& = & f_(1)\lambda\\& = & \text{7.800 s}^(-1) * \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}`

`\begin{array}{rcl}v_(2)& = & f_(2)\lambda\\& = & \text{15.60 s}^(-1) * \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}`

`\begin{array}{rcl}v_(3)& = & f_(3)\lambda\\& = & \text{23.40 s}^(-1) * \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}`