Question 311: Waves

Question

A sonometer wire is stretched by hanging a metal bob. The fundamental frequency of vibration of wire is ' $n_1$ '. When 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

Accepted Answer

The fundamental frequency of vibration of a sonometer wire is determined by the formula:
$ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $
where:
$ n $ is the frequency,
$ L $ is the length of the wire,
$ T $ is the tension in the wire,
$ \mu $ is the mass per unit length of the wire.
When the wire is stretched by hanging a metal bob, the tension $ T $ is due to the weight of the metal bob, thus:
$ T = mg = W $
where $ W $ is the weight of the metal bob.
Given the above formula for frequency and applying it to the initial situation, we have:
$ n_1 = \frac{1}{2L} \sqrt{\frac{W_1}{\mu}} $
When the bob is completely immersed in water, the apparent weight becomes $ W_2 $, giving us:
$ n_2 = \frac{1}{2L} \sqrt{\frac{W_2}{\mu}} $
Therefore, the ratio of the weights before and after immersing in water is:
$ \frac{W_1}{W_2} = \frac{n_1^2}{n_2^2} $
The relative density ($\sigma$) of the bob's material can be calculated as follows:
$ \sigma = \frac{	ext{weight in air}}{	ext{loss of weight in water}} $
Replacing the weights, we get:
$ \frac{W_1}{W_1 - W_2} = \frac{n_1^2}{n_1^2 - n_2^2} $
This equation gives the relative density of the metal of the bob in terms of the frequencies.

Answer

$\frac{\mathrm{n}_1}{\mathrm{n}_1-\mathrm{n}_2}$

Answer

$\frac{\mathrm{H}_2}{\mathrm{n}_1-\mathrm{n}_2}$

Answer

$\mathrm{\frac{n_1^2}{n_1^2-n_2^2}}$

Answer

$\frac{\mathrm{n}_2^2}{\mathrm{n}_1^2-\mathrm{n}_2^2}$

Waves

A sonometer wire is stretched by hanging a metal bob. The fundamental frequency of vibration of wire is ' n_1 '. When 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

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