![The air column in a closed pipe of length 50 cm is set into vibration to generate a wave at its fundamental frequency. i. Sketch the diagram of the wave generated. ii. Calculate I. its wavelength. II. the frequency of the first overtone. [Speed of sound in air = 340 ms-1]](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1783373431320-510ba1718e738492.png&w=3840&q=75)
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here's the solution for part (c): c) i) Sketch the diagram of the wave generated: For a closed pipe at its fundamental frequency, there is a node at the closed end and an antinode at the open end. This corresponds to one-quarter of a wavelength. ` Open End (Antinode) | | /\ | / \ | / \ |/ \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | \ | Node (Closed End) ` ii) Calculate: I. Its wavelength. Step 1: State the given length of the pipe and convert to meters. The length of the closed pipe, L = 50 cm = 0.50 m. Step 2: Relate the pipe length to the wavelength for the fundamental frequency in a closed pipe. For the fundamental frequency in a closed pipe, the length of the pipe is one-quarter of the wavelength (). L = ()/(4) Step 3: Calculate the wavelength. = 4L = 4 × 0.50 m = 2.0 m The wavelength is 2.0 m. II. The frequency of the first overtone. Step 1: State the speed of sound in air. Speed of sound in air, v = 340 m/s. Step 2: Calculate the fundamental frequency (f_1). Using the wave speed formula v = f, we can find the fundamental frequency: f_1 = (v)/() f_1 = 340 m/s2.0 m f_1 = 170 Hz Step 3: Determine the frequency of the first overtone. For a closed pipe, the overtones are odd multiples of the fundamental frequency. The first overtone is the third harmonic (3f_1). f_first overtone = 3 × f_1 f_first overtone = 3 × 170 Hz f_first overtone = 510 Hz The frequency of the first overtone is 510 Hz. What's next?