Superposition of Waves-2 : Stationary (Standing) Waves : Vibration of Air Columns — Practice Quiz
ICSE · Class 11 · Physics
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Quick Quiz: Superposition of Waves-2 : Stationary (Standing) Waves : Vibration of Air Columns
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A closed organ pipe has length 0.80 m. If the speed of sound in air is 320 m/s, what is its fundamental frequency?
An open organ pipe has length 0.75 m and the speed of sound is 330 m/s. What is its second harmonic frequency?
A resonance tube gives first resonance at 0.20 m and second resonance at 0.70 m for a tuning fork of frequency 500 Hz. What is the speed of sound in air?
A resonance tube is used with a fork of frequency 256 Hz. The first and second resonance lengths are 0.32 m and 0.96 m. What is the speed of sound at that temperature?
Sample Questions
A closed pipe has length 0.60 m and radius 0.05 m. If the speed of sound is 340 m/s, what is its fundamental frequency using end correction?
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132.8 Hz
For a closed pipe with end correction, n1 = v / 4(l + 0.6r). Here 0.6r = 0.6 × 0.05 = 0.03 m. Effective length = 0.60 + 0.03 = 0.63 m. So n1 = 340 / (4 × 0.63) = 340 / 2.52 ≈ 132.8 Hz.
A closed pipe has length 0.50 m and speed of sound 340 m/s. What is the difference between its first and second harmonic frequencies?
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340 Hz
For a closed pipe, consecutive harmonics differ by Δn = 2n1 = 2v / 4l. So Δn = 2 × 340 / (4 × 0.50) = 680 / 2 = 340 Hz.
A tuning fork of frequency 256 Hz is used in a resonance tube. If the first resonance length is 0.30 m, what is the second resonance length neglecting end correction?
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0.90 m
For a resonance tube, l1 = λ/4 and l2 = 3λ/4. So l2 = 3l1. If l1 = 0.30 m, then l2 = 0.90 m.
A tuning fork of frequency 400 Hz produces first and second resonance lengths of 0.21 m and 0.63 m in a resonance tube. What is the wavelength of sound?
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0.84 m
Use λ = 2(l2 - l1). Here l2 - l1 = 0.63 - 0.21 = 0.42 m. So λ = 2 × 0.42 = 0.84 m.
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