Question

A metal wire of linear mass density of 9.8 g/m is
stretched with tension of 10 kg weight between two
rigid supports 1 m apart. The fundamental
frequency of the wire is​

Answer 1

Answer:

(a) : At resonance, the frequency of vibration of wire become equal to the frequency of a.c.

Answer 2

The fundamental  frequency of the wire is​ 50 Hz.

Explanation:

The relation between a fundamental frequency and the tension of a wire

is given as

$f=\frac{1}{2L}\sqrt{\frac{T}{M} }$

We can also write,

$f=\frac{1}{2L}\sqrt{\frac{mg}{M} }$

Here f is the fundamental  frequency of the wire, L is the string length and T is the tension in the length and M is the linear mass density.

Given M = 9.8 g/m and L = 1 m and mass, m= 10 kg

Substitute the given values, we get

$f=\frac{1}{2\times1} \sqrt{\frac{10\times9.8}{9.8\times10^-^3} }$

f = 50 Hz

Thus, The fundamental  frequency of the wire is​ 50 Hz.

Learn More: Fundamental  frequency of the wire.

https://brainly.in/question/7000086