Two identical strings of length l and 2l vibrate with fundamental frequencies N Hz and 1.5N Hz, respectively. The ratio of tensions for smaller length to large length is (a) 1:3 (b) 1:9 (c) 3:1 (d) 9:1

Answer: (b). Identical strings (same μ): n = 1/2L√(T/μ) ⇒ T ∝ (nL)². (T₁)/(T₂) = ((N · l)/(1.5N · 2l))² = (1/3)² = 1/9 So the answer is (b) 1:9.

Two identical strings of length l and 2l vibrate with fundamental… | MHT CET 2020 (16 Oct, Shift 1)

Q84MHT CET 2020 · 16 OCT · SHIFT 1Medium

Stationary Waves & Vibrating Strings

Two identical strings of length $l$ and $2l$ vibrate with fundamental frequencies $N\text{ Hz}$ and $1.5N$ Hz, respectively. The ratio of tensions for smaller length to large length is

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Answer: (b)

Identical strings (same $\mu$ ): $n = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}} \Rightarrow T \propto (nL)^2$ .
$\dfrac{T_1}{T_2} = \left(\dfrac{N \cdot l}{1.5N \cdot 2l}\right)^2 = \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9}$
So the answer is (b) $1:9$ .

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