A wire of length 30 cm,… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Waves & Oscillations

2022integerhard

A wire of length $30\mathrm{cm}$ , stretched between rigid supports, has it's ${n}^{\mathrm{th}}$ and ${(n+1)}^{\mathrm{th}}$ harmonics at $400\mathrm{Hz}$ and $450\mathrm{Hz}$ , respectively. If tension in the string is $2700N$ , it's linear mass density is _____ $\mathrm{kg}{m}^{-1}$ .

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Held on 27 Jul 2022 · Verified 6 Jul 2026.

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Solution

Frequency of ${n}^{th}$ harmonic is $\frac{nv}{2l}=400\Rightarrow \frac{nv}{0.6}=400...(1)$

And frequency of ${(n+1)}^{th}$ harmonic is $\frac{(n+1)v}{2l}=450\Rightarrow \frac{(n+1)v}{0.6}=450...(2)$ , here, $v$ is speed of wave on string.

From equation $(1)$ , we have

$n=\frac{400\times 0.6}{v}$

Putting in equation $(2)$ ,

$\Rightarrow [\frac{0.6\times 400}{v}+1]\frac{v}{0.6}=450$

$\Rightarrow [400+\frac{v}{0.6}]=450$

$\Rightarrow v=30m{s}^{-1}$

Speed of wave on string is stated as $v=\sqrt{\frac{T}{\mu }}=30$ , here, $\mu$ is linear mass density.

So, $\mu =\frac{T}{{v}^{2}}=\frac{2700}{900}\Rightarrow \mu =3\mathrm{kg}{m}^{-1}$

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Solution

Frequency of ${n}^{th}$ harmonic is $\frac{nv}{2l}=400\Rightarrow \frac{nv}{0.6}=400...(1)$

And frequency of ${(n+1)}^{th}$ harmonic is $\frac{(n+1)v}{2l}=450\Rightarrow \frac{(n+1)v}{0.6}=450...(2)$ , here, $v$ is speed of wave on string.

From equation $(1)$ , we have

$n=\frac{400\times 0.6}{v}$

Putting in equation $(2)$ ,

$\Rightarrow [\frac{0.6\times 400}{v}+1]\frac{v}{0.6}=450$

$\Rightarrow [400+\frac{v}{0.6}]=450$

$\Rightarrow v=30m{s}^{-1}$

Speed of wave on string is stated as $v=\sqrt{\frac{T}{\mu }}=30$ , here, $\mu$ is linear mass density.

So, $\mu =\frac{T}{{v}^{2}}=\frac{2700}{900}\Rightarrow \mu =3\mathrm{kg}{m}^{-1}$

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Solution

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Solution

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Other Physics topics for JEE Main

Mechanics

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