A=vf-vi/tf-ti
a= velocity final-velocity initial / time final-time initial
a= 6 m/s - 3 m/s divided by 8 s - 0s
a= 2 m/s / 8 m/s
a= 1/4 m/s^2
force times gravity (FG) =mass times gravity (mg)
Answer:
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.