AN ORGAN-GUITAR DUET
A stopped organ pipe is sounded near a guitar, causing one of the strings to vibrate with large amplitude. We vary the string tension until we find the maximum amplitude. The string is 80% as long as the pipe. If both pipe and string vibrate at their fundamental frequency, calculate the ratio of the wave speed on the string to the speed of sound in air.
IDENTIFY and SET UP:
The large response of the string is an example of resonance. It occurs because the organ pipe and the guitar string have the same fundamental frequency. Letting the subscripts a and s stand for the air in the pipe and the string, respectively, the condition for resonance is f_{1a} = f_{1s}. Equation (16.20) gives the fundamental frequency for a stopped pipe, and Eq. (15.32) gives the fundamental frequency for a guitar string held at both ends.
These expressions involve the wave speed in air (v_{a}) and on the string (v_{s}) and the lengths of the pipe and string. We are given that L_{s} = 0.80L_{a}; our target variable is the ratio v_{s}/v_{a}.
f_{1}=\frac{v}{2 L} (15.32)
f_{1}=\frac{v}{4 L} (16.20)
EXECUTE:
From Eqs. (16.20) and (15.32), f_{1a} = v_{a}/4L_{a} and f_{1s} = v_{s}/2L_{s}. These frequencies are equal, so
\frac{v_{\mathrm{a}}}{4 L_{\mathrm{a}}}=\frac{v_{\mathrm{s}}}{2 L_{\mathrm{s}}}Substituting L_{s} = 0.80L_{a} and rearranging, we get v_{s}/v_{a}= 0.40.
EVALUATE: As an example, if the speed of sound in air is 345 m/s, the wave speed on the string is (0.40)(345 m/s) = 138 m/s. Note that while the standing waves in the pipe and on the string have the same frequency, they have different wavelengths λ = v/f because the two media have different wave speeds v. Which standing wave has the greater wavelength?