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: Read the wave number:
Comparing \(y = A\sin(120\pi t + 3x)\) with \(y = A\sin(\omega t + kx)\), the wave number is \(k = 3\) rad/m.

Step 2: Relation:
Phase difference \(\Delta\phi = k\,\Delta x\), so \(\Delta x = \frac{\Delta\phi}{k}\).

Step 3: Calculate:
\[ \Delta x = \frac{\pi/3}{3} = \frac\pi9\text{ m} \]

Final Answer:
The distance is \(\frac\pi9\) m, option (B). \[ \boxed{\frac{\pi}{9}\text{ m}} \]