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.

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

answer: i don't know the answer good luck

explanation: