## College Physics (4th Edition)

(a) The wavelength is $1.2~m$ (b) The next larger value of $L$ is $0.90~m$ (c) $v = 338.4~m/s$
(a) The smallest value of $L$ which produces resonance will be at the fundamental frequency. We can find the wavelength: $\lambda = 4L = (4)(0.30~m) = 1.2~m$ The wavelength is $1.2~m$ (b) The next larger value that will produce resonance will be the third harmonic. We can find the next larger value of $L$: $\frac{4L}{3} = 1.2~m$ $L = \frac{(3)(1.2~m)}{4}$ $L = 0.90~m$ The next larger value of $L$ is $0.90~m$ (c) We can find the speed of sound in the tube: $v = \lambda~f$ $v = (1.2~m)(282~Hz)$ $v = 338.4~m/s$