Question

In a sonometer experiment, the fundamental frequency of vibration of wire is '\(n_1\)' when wire is stretched by hanging a metal bob. If the bob is completely immersed in water, the frequency of vibration of wire becomes '\(n_2\)'. The relative density of the metal of the bob is

• A. \(\frac{n_1}{n_1-n_2}\)
• B. \(\frac{n_1-n_2}{n_2}\)
• C. \(\frac{n_1^2}{n_1^2-n_2^2}\)
• D. \(\frac{n_2^2}{n_1^2-n_2^2}\)

Hint:

Phase difference equals wave number times path difference.

Answer 1

The Correct Option is C

Step 1: Wavelength route:
$\lambda = \frac{2\pi}{k} = \frac{2\pi}{3}$ m.

Step 2: Fraction of a wavelength:
A phase difference of $\frac\pi3$ is $\frac16$ of $2\pi$, so the distance is $\frac{1}{6}\times\frac{2\pi}{3} = \frac\pi9$ m (B).

Final Answer:
$\frac\pi9$ m. \[ \boxed{\frac{\pi}{9}\text{ m}} \]