(a) What should be the length of an organ pipe, closed at one end, if the fundamental frequency is to be \(261.5 \mathrm{Hz} ?\) (b) What is the fundamental frequency of the organ pipe of part (a) if the temperature drops to \(0.0^{\circ} \mathrm{C} ?\)

To find the length of an organ pipe closed at one end, we can use the formula for the fundamental frequency of a closed organ pipe: $$f_1 = \frac{v}{4L}$$ Here, \(f_1\) is the fundamental frequency, \(v\) is the speed of sound in air, and \(L\) is the length of the organ pipe. We also need to know the speed of sound in air at a given temperature. The formula for the speed of sound in air is: $$v = 331.3 \sqrt{1 + \frac{T}{273.15}}$$ where \(T\) is the temperature in Celsius. Assuming room temperature (\(20°C\)), we can find the speed of sound in air: $$v = 331.3 \sqrt{1 + \frac{20}{273.15}} \approx 343 m/s$$ Now we can calculate the length of the organ pipe by rearranging the formula for the fundamental frequency: $$L = \frac{v}{4f_1}$$ Plugging in the given fundamental frequency \(f_1 = 261.5 \mathrm{Hz}\) and the calculated speed of sound \(v \approx 343 m/s\): $$L = \frac{343}{4(261.5)} \approx 0.328 m$$ So the length of the organ pipe should be approximately \(0.328 m\).

To find the fundamental frequency of the organ pipe of part (a) when the temperature drops to \(0.0^{\circ} \mathrm{C}\), we first need to find the speed of sound in air at this temperature. Using the same formula for the speed of sound in air: $$v' = 331.3 \sqrt{1 + \frac{0}{273.15}} \approx 331.3 m/s$$ Now, we can use the fundamental frequency formula for a closed organ pipe as before: $$f'_1 = \frac{v'}{4L}$$ Plugging in the calculated speed of sound at 0°C, \(v' \approx 331.3 m/s\), and the length of the organ pipe from part (a), \(L \approx 0.328 m\): $$f'_1 = \frac{331.3}{4(0.328)} \approx 252.4 \mathrm{Hz}$$ So the fundamental frequency of the organ pipe of part (a) when the temperature drops to \(0.0^{\circ} \mathrm{C}\) is approximately \(252.4 \mathrm{Hz}\).