A steel wire in a piano (0.7000m) (4300g).what tension must this wire stretched so that fundamental vibration with frequency of 261.6Hz?

1 Answer
Sep 13, 2015

#T=8.24xx10^5"N"#

Explanation:

www.schoolphysics.co.uk

We are interested in the 1st diagram which refers to the fundamental frequency. It can be shown that:

#f=(1)/(2L)sqrt((T)/(m))#

#f# = fundamental frequency

#L# = length

#T# = tension

#m# = mass per unit length.

#sqrt((T)/(m))=2Lf#

#(T)/(m)=4f^2L^2#

#T=4f^2L^2m#

#T=4xx261.6^2xx0.7^2xx4.3/0.7#

#T=8.24xx10^5"N"#