JEE Advanced 2023 — Paper 2 — PHY-III solution

JEE Advanced2023Paper 2PHY-III

A string of length $1$ m and mass $2 \times 1 0^{- 5}$ kg is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies $750$ Hz and $1000$ Hz. The value of tension $T$ is _______ Newton.

Your answer

Solution

For a string fixed at both ends with linear mass density $μ$ , the $p$ -th harmonic has frequency

$f_{p} = \frac{p}{2 L} \frac{T}{μ}$ .

Two successive harmonics, $p$ and $p + 1$ , satisfy

$\frac{f _{p + 1}}{f _{p}} = \frac{p + 1}{p} = \frac{1000}{750} = \frac{4}{3} \Rightarrow p = 3$ .

Therefore from $f_{3} = 750$ Hz with $L = 1$ m,

$750 = \frac{3}{2} \frac{T}{μ} \Rightarrow \frac{T}{μ} = 500$ .

With $μ = 2 \times 1 0^{- 5}$ kg m $^{- 1}$ ,

$T = 50 0^{2} \cdot 2 \times 1 0^{- 5} = 25 \times 1 0^{4} \cdot 2 \times 1 0^{- 5} = 5 N$ .

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