91Ó°ÊÓ

We are given the mass \(m\) and length \(L\) of the string. We can find the linear mass density \(\mu\) given by: \(\mu = \frac{m}{L}\). In our case, we have the mass in grams, so we need to convert it to kg: $$ \mu = \frac{0.230 \times 10^{-3} \,\text{kg}}{30.0 \times 10^{-2} \,\text{m}} = 7.67 \times 10^{-5} \,\text{kg/m} $$

We know that the string is vibrating at its next-to-lowest natural frequency \(f_{2}\). We have to first find the fundamental frequency \(f_1\) using the following equation: $$ f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $$ where \(T\) is the tension in the string, and \(L\) is the length of the string. Plugging in the values: $$ f_1 = \frac{1}{2 \times 30.0 \times 10^{-2} \, \text{m}}\sqrt{\frac{7.00 \,\text{N}}{7.67 \times 10^{-5} \,\text{kg/m}}} = 85.8\,\text{Hz} $$

Since the string is vibrating at the next-to-lowest natural frequency \(f_{2}\), which corresponds to the second harmonic, we can find by doubling the fundamental frequency \(f_1\): $$ f_2 = 2 \times f_1 = 2 \times 85.8\,\text{Hz} = 171.6\,\text{Hz} $$ So, the next-to-lowest natural frequency \(f_{2}\) is \(171.6\,\text{Hz}\).

Now that we have the frequency \(f_{2}\), we will use it to find the frequency and wavelength of the sound in the air. The frequency of the sound wave in the air will be the same as the frequency of the string's vibration: $$ f_\text{air} = f_2 = 171.6\,\text{Hz} $$ Next, we need to find the wavelength of the sound in the air. We know the speed of sound \(v_\text{air} = 350\,\text{m/s}\) and we found the frequency. We can find the wavelength using the following equation: $$ v_\text{air} = f_\text{air} \times \lambda_\text{air} $$ We can solve for \(\lambda_\text{air}\): $$ \lambda_\text{air} = \frac{v_\text{air}}{f_\text{air}} = \frac{350\,\text{m/s}}{171.6\,\text{Hz}} = 2.04\,\text{m} $$ Thus, the frequency of the sound in the air is \(171.6\,\text{Hz}\), and the wavelength is \(2.04\,\text{m}\).