A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s).
For pipe closed at one end:
Given: L = 20 cm = 0.2 m, v = 340 m/s, source frequency = 430 Hz
For a pipe closed at one end, the harmonic frequencies are: v_n = nv/4L where n = 1, 3, 5, ... (odd harmonics only)
Fundamental frequency: v₁ = v/4L = 340/() = 340/0.8 = 425 Hz
Since the source frequency (430 Hz) is very close to the fundamental frequency (425 Hz), the first harmonic (fundamental mode) is resonantly excited.
For pipe open at both ends:
For an open pipe, the harmonic frequencies are: v_n = nv/2L where n = 1, 2, 3, ...
Fundamental frequency: v₁ = v/2L = 340/() = 340/0.4 = 850 Hz
Since 430 Hz does not match any harmonic frequency of the open pipe (the fundamental itself is 850 Hz), no resonance will be observed with the same source when both ends are open.
Explanation
This question tests understanding of resonance in closed and open pipes. For a closed pipe, only odd harmonics exist with frequencies nv/4L. The calculated fundamental is 425 Hz, close to 430 Hz, so resonance occurs at the first harmonic. For an open pipe, all harmonics exist with frequencies nv/2L. The fundamental is 850 Hz, and since 430 Hz doesn't match any harmonic (n would need to be 0.506, not an integer), no resonance occurs.
Solution Steps
Step 1: Convert length to metres: L = 20 cm = 0.2 m
Step 2: For closed pipe, calculate fundamental frequency: v₁ = v/4L = 340/() = 425 Hz
Step 3: Compare source frequency (430 Hz) with fundamental (425 Hz) - they match approximately, so 1st harmonic is excited
Step 4: For open pipe, calculate fundamental frequency: v₁ = v/2L = 340/() = 850 Hz
Step 5: Check if 430 Hz matches any harmonic: n = L/v = 0.506 (not an integer), so no resonance