IBDP Physics- C.4 Standing waves and resonance- IB Style Questions For HL Paper 1A -FA 2025
▶️ Answer/Explanation
For a pipe closed at one end: L = (2n-1)λ/4 where n = 1,2,3,…
Fifth harmonic means n = 5: L = (2×5-1)λ/4 = 9λ/4 ⇒ λ = 4L/9
But 4L/9 is not an option. Let’s check: fundamental (n=1): L = λ/4
Third harmonic (n=2): L = 3λ/4
Fifth harmonic (n=3): L = 5λ/4 ⇒ λ = 4L/5
✅ Answer: (C) 4L/5
▶️ Answer / Explanation
For a pipe that is closed at one end and open at the other, a node is formed at the closed end and an antinode at the open end.
Only odd harmonics are present in such a pipe.
The fundamental frequency is:
\( f_1 = \dfrac{V}{4L} \)
The general expression for the frequencies of the harmonics is:
\( f_n = \dfrac{(2n – 1)V}{4L} \quad \text{where } n = 1,2,3,\dots \)
This correctly excludes even harmonics.
✅ Answer: (B)
▶️ Answer / Explanation
For a pipe that is closed at one end and open at the other, only odd harmonics are present.
The allowed harmonic frequencies are given by:
\( f_n = (2n – 1)f_1 \quad \text{where } n = 1,2,3,\dots \)
The third harmonic corresponds to:
\( f_3 = 3f_1 \)
Given \( f_3 = 150\,\mathrm{Hz} \),
\( f_1 = \dfrac{150}{3} = 50\,\mathrm{Hz} \)
Therefore, another possible standing-wave frequency in this pipe is the fundamental frequency of \(50\,\mathrm{Hz}\).
✅ Answer: (B)