Two wires made of the same material have same length and the tension.Their diameters are 1 mm (wire 1)and 0.5 mm (wire 2).The ratio f₂/f₁ of their fundamental frequencies is A)0.10. B)0.25. C)0.5. D)1.0. E)2.0

Question

Quizplus · Accepted Answer

The fundamental frequency of a string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density ($\mu = \frac{m}{L}$, with $m$ being the mass and $L$ the length of the wire). The linear mass density is directly proportional to the cross-sectional area of the wire ($\mu \propto A$), and the area $A$ is proportional to the square of the diameter ($A \propto d^2$). Therefore, if the diameter of wire 2 is half that of wire 1 ($d_2 = \frac{1}{2}d_1$), then its area, and thus its linear mass density, is a quarter ($\mu_2 = \frac{1}{4}\mu_1$). Since the fundamental frequency is inversely proportional to the square root of the linear mass density ($f \propto \frac{1}{\sqrt{\mu}}$), the ratio of the fundamental frequencies of wire 2 to wire 1 is $\sqrt{4} = 2$. This means the fundamental frequency of wire 2 is twice that of wire 1.

Two Wires Made of the Same Material Have Same Length