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# A string oscillating at a fundamental frequency under a tension of $225\;{\text{N}}$ produces $6\;{\text{beats/sec}}$ with a sonometer. If the tension is $256\;{\text{N}}$, then again oscillating at fundamental note it produces $6$ beats per second with the same sonometer. The frequency of the sonometer is:A) $256\;{\text{Hz}}$ B) $225\;{\text{Hz}}$C) $280\;{\text{Hz}}$D) $186\;{\text{Hz}}$

Last updated date: 13th Aug 2024
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Hint: The difference in frequencies of two waves can be termed as beats per second. The frequency increases when the tension increases. The fundamental frequency of the sonometer can be found by comparing the two fundamental frequencies of different tensions.

The expression for the fundamental frequency of the for a string is given as,
$\eta = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{\mu }}$
Where, $T$ is tension, $\mu$ is the linear mass density and $l$ is the length of the string.
Let ${\eta _1}$ is the fundamental frequency of the string under a tension of $225\;{\text{N}}$.
Therefore,
${\eta _1} = \dfrac{1}{{2l}}\sqrt {\dfrac{{225}}{\mu }} \\ = \dfrac{{15}}{{2l}}\sqrt {\dfrac{1}{\mu }} \\$
Let ${\eta _1}^\prime$ is the fundamental frequency of the string under a tension of $256\;{\text{N}}$.
Therefore,
${\eta _1}^\prime = \dfrac{1}{{2l}}\sqrt {\dfrac{{256}}{\mu }} \\ = \dfrac{{16}}{{2l}}\sqrt {\dfrac{1}{\mu }} \\$
And let the fundamental frequency of the sonometer is ${\eta _2}$ .
Also let’s take $k = \dfrac{1}{{2l}}\sqrt {\dfrac{1}{\mu }}$
Therefore,${\eta _1} = 15k$ and ${\eta _1}^\prime = 16k$.
The difference in fundamental frequencies can be termed as beats per second. It is given that string oscillating at a fundamental frequency under a tension of $225\;{\text{N}}$ produces $6\;{\text{beats/sec}}$ with a sonometer.
Therefore, ${\eta _2} - {\eta _1} = 6........\left( 1 \right)$
Also it is given that string oscillating at a fundamental frequency under a tension of $256\;{\text{N}}$ produces $6\;{\text{beats/sec}}$ with a sonometer.
Therefore, ${\eta _1}^\prime - {\eta _2} = 6........\left( 2 \right)$
Solving equation $\left( 1 \right)$ and equation $\left( 2 \right)$, we get
${\eta _1}^\prime - {\eta _1} = 12$
Substituting for the above expression,
$16k - 15k = 12 \\ k = 12 \\$
From the equation $\left( 1 \right)$ and above results,
${\eta _2} - {\eta _1} = 6 \\ {\eta _2} = {\eta _1} + 6 \\ {\eta _2} = 15k + 6 \\$
Substitute the values, we get
${\eta _2} = 15 \times 12 + 6 \\ = 186\;{\text{Hz}} \\$
Thus the fundamental frequency of the sonometer is $186\;{\text{Hz}}$.