Question

What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.630 m long and has a mass of 5.69 g.

Answer 1

Answer:

220.698Hz

Explanation:

The fundamental frequency f₀ is expressed as f₀ =V/2L where;

V is the speed of the string = $\sqrt{\frac{T}{M} }$

m is the mass of the string

L is the length of the string

T is the tension in the string

f₀ = $\frac{1}{2L} \sqrt{\frac{T}{m} }$

Given datas

m = 5.69g = 0.00569 kg

T = 440N

L = 0.630 m

Required

Fundamental frequency of the steel piano wire f₀

$f_0 = \frac{1}{2(0.630)}\sqrt{\frac{440}{0.00569} } \\ \\f_0 = \frac{1}{1.26}\sqrt{77,328.65 } \\\\f_0 = \frac{1}{1.26} * 278.08\\\\f_0 = 220.698Hz$

Hence the frequency of the fundamental mode of vibration of the steel piano wire stretched to a tension of 440N is 220.698Hz