A long tube contains air at a pressure of 1.00 atm and a temperature of 77.0°C. The tube is open at one end and closed at the other by a movable piston. A tuning fork that vibrates with a frequency of 500 Hz is placed near the open end. Resonance is produced when the piston is at distances 18.0 cm, 55.5 cm, and 93.0 cm from the open end. (a) From these values, what is the speed of sound in air at 77.0°C? (b) From the result of part (a), what is the value of g? (c) These results show that a displacement antinode is slightly outside the open end of the tube. How far outside is it?

The length of the pipe is $L=\frac{n\lambda }{2}$,the velocity of the sound is $v=f\lambda$and the relations of the speed of sound v androle="math" localid="1664352505587" $\gamma$is$v=\sqrt{\frac{\gamma RT}{M}}$

The length of the pipe is calculated by L = 93 cm - 55.5 cm = 37.5. The length of the pipe for one node is related to the wavelength of the wave and it is given by equation (16.17) in the form $L=\frac{n\lambda }{2}$.Substitute the values in equation $L=\frac{n\lambda }{2}$we get,

$\lambda =\frac{2L}{n}=\frac{2(37.5)}{1}=75cm$

The velocity of the sound wave equals the time product of the frequency and the wavelength $\mathrm{v}=\mathrm{f\lambda }$. Substitute the values we get,

$\mathrm{v}=\mathrm{f\lambda }=\left(500\mathrm{Hz}\right)(0.75\mathrm{m})=375\mathrm{m}/\mathrm{s}$

Therefore, the value of v=375m/s

The equation that states the relation between the speed of sound v and -y is equation (16.10) in the form $v=\sqrt{\frac{\gamma RT}{M}}$Where is the ratio of heat capacities, M is the molar mass and R is gas constant. Solve this equation for$\gamma$

$$\begin{gathered} \gamma =\frac{M{v}^2}{RT} \\ =\frac{(28.8\times {10}^{-3}(375)}{(8.314)\left(350\right)} \\ =1.40 \end{gathered}$$

Therefore, the value of$\gamma =1.40$

The length of one node is given by $\mathrm{l}=\frac{\mathrm{\lambda }}{4}=\frac{75}{4}=18.75\mathrm{cm}$ The resonance is produced when the piston is at distances 18 cm, so the change in displacement is calculate$\mathrm{\Delta l}=18.75\mathrm{cm}-18\mathrm{cm}=0.75\mathrm{cm}$

Therefore, the resonance is given by 0.75cm