A string under a tension of 129.6,, N produces 10,, beats /sec when it is vibrated along with

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14.Waves and Sound

hard

A string under a tension of $129.6\,\, N$ produces $10\,\, beats /sec$ when it is vibrated along with a tuning fork. When the tension is the string is increased to $160\,\, N,$ it sounds in unison with same tuning fork. calculate fundamental freq. of tuning fork .... $Hz$

$100$

$50$

$150$

$200$

Solution

frequency of tuning fork $ = \frac{{\sqrt {\frac{{160}}{{\rm{m}}}} }}{{2l}}$

According to the question

$\frac{1}{{2l}}\sqrt {\frac{{160}}{{\rm{m}}}} – \frac{1}{{2l}}\sqrt {\frac{{129.6}}{{\rm{m}}}} = 10$

$\frac{1}{{2l}}\sqrt {\frac{{10}}{m}} [4 – 3.6] = 10 \Rightarrow \frac{1}{{2l}}\sqrt {\frac{{10}}{m}} = 25$

so ${{\rm{V}}_{{\rm{TF}}}} = \frac{1}{{2l}}\sqrt {\frac{{160}}{{\rm{m}}}} = 4\left[ {\frac{1}{{2l}}\sqrt {\frac{{10}}{{\rm{m}}}} } \right] = 100\,{\rm{Hz}}$

Standard 11

Physics

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easy

A string under a tension of $129.6\,\, N$ produces $10\,\, beats /sec$ when it is vibrated along with a tuning fork. When the tension is the string is increased to $160\,\, N,$ it sounds in unison with same tuning fork. calculate fundamental freq. of tuning fork .... $Hz$

Solution

Similar Questions

Explain the reflection of wave at rigid support.

A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire ?

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