Question

If the tension and diameter of a sonometer wire of fundamental frequency n is doubled and density is halved then its fundamental frequency will become

• A. $\frac{n}{4}$
• B. $\sqrt{2n}$
• C. n
• D. $\frac{n}{\sqrt{2}}$

Answer 1

Answer: (C)

n

(c) $n=\frac{1}{2l}\sqrt{\frac{T}{\pi r^2\rho }}\propto \sqrt{\frac{T}{r^2\rho }}$

$\Rightarrow \frac{n_1}{n_2}=\sqrt{\left(\frac{T_1}{T_2}\right){\left(\frac{r_2}{r_1}\right)}^2\left(\frac{{\rho }_2}{{\rho }_1}\right)}=\sqrt{\left(\frac{1}{2}\right){\left(\frac{2}{1}\right)}^2\left(\frac{1}{2}\right)}=1$

$\therefore n_1=n_2=n$