An open organ pipe of length ' l ' is sounded together with another open organ pipe of length (l + l₁) in their fundamental modes. Speed of sound in air is ' V '. The beat frequency heard will be ( l₁ << l) (a) (Vl₁²)/(2l) (b) (Vl₁)/(2l²) (c) (Vl₁)/(4l²) (d) (vl²)/(2l₁)

Answer: (b). Open pipe fundamental: f = V/2l. Beats = V/2l - (V)/(2(l+l₁)) = (Vl₁)/(2l(l+l₁)) Since l₁ l, l(l+l₁) ≈ l², so beats ≈ (Vl₁)/(2l²). So the answer is (b) (Vl₁)/(2l²).

An open organ pipe of length ' l ' is sounded together with another… | MHT CET 2024 (15 May, Shift 1)

Q37MHT CET 2024 · 15 MAY · SHIFT 1Medium

Beats

An open organ pipe of length ' $l$ ' is sounded together with another open organ pipe of length $(l + l_1)$ in their fundamental modes. Speed of sound in air is ' $V$ '. The beat frequency heard will be ( $l_1 << l)$

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Answer: (b)

Open pipe fundamental: $f = \frac{V}{2l}$ .
Beats $= \frac{V}{2l} - \frac{V}{2(l+l_1)} = \frac{Vl_1}{2l(l+l_1)}$
Since $l_1 \ll l$ , $l(l+l_1) \approx l^2$ , so beats $\approx \frac{Vl_1}{2l^2}$ .
So the answer is (b) $\frac{Vl_1}{2l^2}$ .

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Beats question from MHT CET 2024 (15 May, Shift 1)

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