Apiano tuner stretches a steel piano wire with a tension of 800 n. the steel wire is 0.400 m long and has a mass of 3.00 g. what is the frequency of its fundamental mode of vibration?

408.25 Hz.

Explanation:

The fundamental frequency of a stretched string is given as

f' = 1/2L√(T/m') Equation 1

Note: The a steel piano wire is a string

Where f' = fundamental frequency of the wire, L = length of the wire, T = tension on the wire, m' = mass per unit length of the wire.

Given: L = 0.4 m, T = 800 N,

Also,

m' = m/L where m = mass of the steel wire = 3.00 g = 3/1000 = 0.003 kg.

L = 0.4 m

m' = 0.003/0.4 = 0.0075 kg/m.

Substituting into equation 1

f' = 1/(2×0.4)[√(800/0.0075)]

f' = 1/0.8[√(106666.67)]

f' = 326.599/0.8

f' = 408.25 Hz.

Hence the frequency of the fundamental mode of vibration = 408.25 Hz.

he . if i may ask what i your question exactly?

explanation:

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!