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If the tension and diameter of a sonometer wire of fundamental frequency n are doubled and density is halved then its fundamental frequency will become
A
$\dfrac{n}{4}$
B
$\sqrt{2}n$
C
n
D
$\dfrac{n}{\sqrt{2}}$
Detailed Solution
Fundamental frequency of a stretched wire: $n = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$, where the mass per unit length $\mu = \pi r^2\rho$.
$n = \dfrac{1}{2l}\sqrt{\dfrac{T}{\pi r^2\rho}}$
New values: $T_1 = 2T$, diameter doubled so $r_1 = 2r$, $\rho_1 = \dfrac{\rho}{2}$
$n_1 = \dfrac{1}{2l}\sqrt{\dfrac{2T}{\pi(2r)^2\frac{\rho}{2}}} = \dfrac{1}{2l}\sqrt{\dfrac{2T}{2\pi r^2\rho}} = \dfrac{1}{2l}\sqrt{\dfrac{T}{\pi r^2\rho}} = n$
So there is no change: the frequency remains n.
$n = \dfrac{1}{2l}\sqrt{\dfrac{T}{\pi r^2\rho}}$
New values: $T_1 = 2T$, diameter doubled so $r_1 = 2r$, $\rho_1 = \dfrac{\rho}{2}$
$n_1 = \dfrac{1}{2l}\sqrt{\dfrac{2T}{\pi(2r)^2\frac{\rho}{2}}} = \dfrac{1}{2l}\sqrt{\dfrac{2T}{2\pi r^2\rho}} = \dfrac{1}{2l}\sqrt{\dfrac{T}{\pi r^2\rho}} = n$
So there is no change: the frequency remains n.