Question 380: Waves

Question

A sonometer wire '$$A$$' of diameter '$$\mathrm{d}$$' under tension '$$T$$' having density '$$\rho_1$$' vibrates with fundamental frequency '$$n$$'. If we use another wire '$$B$$' which vibrates with same frequency under tension '$$2 \mathrm{~T}$$' and diameter '$$2 \mathrm{D}$$' then density '$$\rho_2$$' of wire '$$B$$' will be

Accepted Answer

The formula for frequency of a sonometer is $$\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi ho \mathrm{D}^2}}$$
Here $$l$$ is length, T is tension, D is diameter and $$ho$$ is density.
The frequency of both the wires is same.
The frequency of the wire $$A$$ is $$f_A=\frac{1}{2 l} \sqrt{\frac{T}{\pi ho_1 D^2}}$$
The frequency of the wire $$B$$ is $$f_B=\frac{1}{2 l} \sqrt{\frac{2 T}{\pi ho_2(2 D)^2}}$$
Equating both the frequencies
$$\begin{aligned}
& \frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi ho \mathrm{D}^2}}=\frac{1}{2 l} \sqrt{\frac{2 \mathrm{~T}}{\pi ho_2(2 \mathrm{D})^2}} \
& \sqrt{\frac{1}{ho_1}}=\sqrt{\frac{1}{2 ho_2}} \
& \frac{1}{ho_1}=\frac{1}{2 ho_2} \
	herefore \quad & ho_2=\frac{ho_1}{2}
\end{aligned}$$

Answer

$$\rho_2=2 \rho_1$$

Answer

$$\rho_2=\rho_1$$

Answer

$$\rho_2=\frac{\rho_1}{2}$$

Answer

$$\rho_2=\frac{\rho_1}{4}$$

Waves

A sonometer wire '$A' of diameter '\mathrm{d}' under tension 'T' having density '\rho_1' vibrates with fundamental frequency 'n'. If we use another wire 'B' which vibrates with same frequency under tension '2 \mathrm{~T}' and diameter '2 \mathrm{D}' then density '\rho_2' of wire 'B$' will be

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