Question 411: Waves

Question

A sonometer wire under suitable tension having specific gravity $$\rho$$, vibrates with frequency $$n$$ in air. If the load is completely immersed in water the frequency of vibration of wire will become

Accepted Answer

The fundamental frequency of a sonometer wire is given by:
$ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $
Here, $ L $ (the length) and $ \mu $ (the linear mass density) are constants, so the frequency $ n $ is dependent on the tension $ T $.
We have:
$ n \propto \sqrt{T} $
When considering the wire's vibration after submerging the load in water, the change in frequency can be determined by comparing the tension in both scenarios (before and after immersion):
$ \frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} $
Where:
$ (	ext{Weight})_s = 	ext{Weight of the object in air} = V \cdot ho_s \cdot g $
$ (	ext{Weight})_w = 	ext{Weight of the object in water} = V \cdot ho_w \cdot g $
Thus:
$ \frac{n_2}{n_1} = \sqrt{\frac{V \cdot ho_s \cdot g - V \cdot ho_w \cdot g}{V \cdot ho_s \cdot g}} = \sqrt{\frac{\frac{ho_s}{ho_w} - 1}{\frac{ho_s}{ho_w}}} = \sqrt{\frac{ho - 1}{ho}} $
Given that the specific gravity of the sonometer wire is:
$ 	ext{Specific gravity, } ho = \frac{ho_s}{ho_w} $
The frequency $ n_2 $ after submerging the load in water is:
$ n_2 = n \sqrt{\frac{ho - 1}{ho}} $

Answer

$$\left[\frac{\rho-1}{n \rho}\right]^{\frac{1}{2}}$$

Answer

$$n\left[\frac{\rho-1}{\rho}\right]^{\frac{1}{2}}$$

Answer

$$n\left[\frac{\rho}{\rho-1}\right]^{\frac{1}{2}}$$

Answer

$$\left[\frac{n \rho}{\rho-1}\right]^{\frac{1}{2}}$$

Waves

A sonometer wire under suitable tension having specific gravity $\rho, vibrates with frequency n$ in air. If the load is completely immersed in water the frequency of vibration of wire will become

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