Question

In a sonometer experiment, the bridges are separated by a fixed distance. The wire which is slightly elastic emits a tone of tension is increased to $4T$, the tone emitted by the wire will be of a frequency
(A) $n$
(B) $2n$
(C) slightly greater than $2n$
(D) slightly less than $2n$

Answer 1

Hint: In sonometer experiment, the bridges are separated by a fixed distance. And the tension is changed, then the tone emitted by the wire will be the frequency of $n$. Then the frequency can be determined by using the formula of vibration of the string.

Formula used:
Vibration of the string or the frequency o0f the string is given by,
$n = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{\mu }} $
Where $L$ is the length of the string, $T$ is the tension of the spring, and $\mu $ is the linear density(mass per unit length).

Complete step by step answer:
Given that,
In a sonometer experiment, the bridges are separated by a fixed distance. The wire which is slightly elastic emits a tone of tension that is increased to $4T$.
Now,
Vibration of the string or the frequency o0f the string is given by,
$n = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{\mu }} \,........................\left( 1 \right)$
In the above equation (1), the linear density (mass per unit length) is written as,
$\Rightarrow n = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{{\pi {r^2}\rho }}} \,....................\left( 2 \right)$
Where, $r$ is the radius of the string and $\rho $ is the density of the string.
If the tension is increased from $T$ to $4T$, then the above equation is written as,
$\Rightarrow {n^{'}} = \dfrac{1}{{2{L^{'}}}}\sqrt {\dfrac{{4T}}{{\pi {r^{'}}^2\rho }}} \,.......................\left( 3 \right)$
By dividing the equation (3) and equation (2), then
$\Rightarrow \dfrac{{{n^{'}}}}{n} = \dfrac{{\dfrac{1}{{2L}}\sqrt {\dfrac{{4T}}{{\pi {r^2}\rho }}} }}{{\dfrac{1}{{2{L^{'}}}}\sqrt {\dfrac{T}{{\pi {r^{{'}2}}\rho }}} }}$
By cancelling the same terms in the above equation, then the above equation is written as,
$\Rightarrow \dfrac{{{n^{'}}}}{n} = \dfrac{{\dfrac{1}{L}\sqrt {\dfrac{4}{{{r^{{'}2}}}}} }}{{\dfrac{1}{{{L^{'}}}}\sqrt {\Rightarrow \dfrac{1}{{{r^2}}}} }}$
By rearranging the terms, then
$\Rightarrow \dfrac{{{n^{'}}}}{n} = 2\left( {\dfrac{{{L^{'}}}}{L} \times \dfrac{{{r^{'}}}}{r}} \right)$
From the equation (1), it shows that
$\Rightarrow n \propto \sqrt T $
If the tension is increased from $T$ to $4T$, then the above equation is written as,
$\Rightarrow n \propto \sqrt {4T} $
By taking root, then
$\Rightarrow n \propto 2\sqrt T $
Then the above equation is written as,
$\Rightarrow {n^{'}} \propto 2n$

But the length is inversely proportional to the frequency, when the tension is increased, the frequency is less than $2n$. Hence, the option (D) is the correct answer.

Note:
In Physics, it is used to measure the tension, frequency, or density of vibrations. The vibrations produced by the string works under the principle of resonance and is often represented as sinusoidal waves. A Sonometer is a device for demonstrating the relationship between the frequency of the sound produced by a plucked string, and the tension, length, and mass per unit length of the string. For small-amplitude vibration, the frequency is proportional to the square root of the tension of the string.