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Waves question

2024 · 1 Feb · Shift 1 · Q89
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Waves question

2024 · 1 Feb · Shift 1 · Q89

JEE MainPhysicsWavesNumerical+4 / −1
A tuning fork resonates with a sonometer wire of length 1 m1 \mathrm{~m}1 m stretched with a tension of 6 N6 \mathrm{~N}6 N. When the tension in the wire is changed to 54 N54 \mathrm{~N}54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is ‾Hz\underline{\hspace{2cm}}\mathrm{Hz}​Hz.
Numerical answer
View written solutionFree

Correct answer: 6

  1. Frequency of a sonometer wire

For a given wire vibrating in the fundamental mode,

f=12LTμf=\frac{1}{2L}\sqrt{\frac{T}{\mu}}f=2L1​μT​​

Since the same wire and same length are used, we have

f∝Tf \propto \sqrt{T}f∝T​

  1. At tension T1=6 NT_1=6\,\text{N}T1​=6N

The wire resonates with the tuning fork, so the wire frequency equals the fork frequency.

Let the tuning fork frequency be fff.

Thus, at 6 N6\,\text{N}6N,

f1=ff_1=ff1​=f

  1. At tension T2=54 NT_2=54\,\text{N}T2​=54N

Using proportionality,

f2f1=T2T1=546=9=3\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{54}{6}}=\sqrt{9}=3f1​f2​​=T1​T2​​​=654​​=9​=3

So,

f2=3ff_2=3ff2​=3f

  1. Beat condition

When the tension is changed to 54 N54\,\text{N}54N, the beat frequency with the same tuning fork is 12 Hz12\,\text{Hz}12Hz.

Beat frequency is

∣f2−f∣=12|f_2-f|=12∣f2​−f∣=12

Substitute f2=3ff_2=3ff2​=3f:

∣3f−f∣=12|3f-f|=12∣3f−f∣=12 2f=122f=122f=12 f=6 Hzf=6\,\text{Hz}f=6Hz

  1. Final answer

The frequency of the tuning fork is

6 Hz\boxed{6\,\text{Hz}}6Hz​

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