When a string of length ' l ' is divided into three segments of length l₁, l₂ and l₃. The fundamental frequencies of three segments are n₁, n₂ and n₃ respectively. The original fundamental frequency ' n ' of the string is (a) n = n₁ + n₂ + n₃ (b) √(n) = √(n₁) + √(n₂) + √(n₃) (c) 1/n = 1/n₁ + 1/n₂ + 1/n₃ (d) 1/√(n) = (1)/(√(n₁)) + (1)/(√(n₂)) + (1)/(√(n₃))

Answer: (c). Same tension and μ: n = k/l, so l = k/n. l = l₁ + l₂ + l₃ ⇒ k/n = k/n₁ + k/n₂ + k/n₃. 1/n = 1/n₁ + 1/n₂ + 1/n₃. So the answer is (c).

When a string of length ' l ' is divided into three segments of… | MHT CET 2023 (9 May, Shift 1)

Q61MHT CET 2023 · 9 MAY · SHIFT 1Easy

Stationary Waves & Vibrating Strings

When a string of length ' $l$ ' is divided into three segments of length $l_1, l_2$ and $l_3$ . The fundamental frequencies of three segments are $n_1, n_2$ and $n_3$ respectively. The original fundamental frequency ' $n$ ' of the string is

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Answer: (c)

Same tension and $\mu$ : $n = \dfrac{k}{l}$ , so $l = \dfrac{k}{n}$ .
$l = l_1 + l_2 + l_3 \Rightarrow \dfrac{k}{n} = \dfrac{k}{n_1} + \dfrac{k}{n_2} + \dfrac{k}{n_3}$ .
$\dfrac{1}{n} = \dfrac{1}{n_1} + \dfrac{1}{n_2} + \dfrac{1}{n_3}$ .
So the answer is (c).

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