Question

A guitar string has a mass per length of 2.33×10−3kg/m and a fundamental frequency of 146.8 Hz when it is under a tension of 82.4 N. The string breaks, and its owner has only a spare string of mass per length 6.80×10−3kg/m.
What tension should the 6.80×10−3kg/m string have so that its fundamental frequency is 146.8 Hz?

Answer 1

Answer:

240.59 N

Explanation:

We are given that

Mass per unit length=m/l=$\mu=2.33\times 10^{-3} kg/m$

Fundamental frequency,f=146.8 Hz

Tension=T=82.4 N

Mass per unit length=$\mu'=6.8\times 10^{-3} kg/m$

We have to find the tension.

Velocity,v=$\sqrt{\frac{T}{\mu}}=\sqrt{\frac{82.4}{2.33\times 10^{-3}}}=188.1m/s$

New tension,T'=$v^2\mu'=(188.1)^2\times 6.8\times 10^{-3}$

$T'=240.59 N$

Hence, the string  should have tension 240.59 N  so that its fundamental frequency is 146.8 Hz

Answer 2

Answer:

Explanation:

mass per unit length, μ = 2.33 x 10^-3 kg/m

frequency, f = 146.8 Hz

Tension, T = 82.4 N

mass per unit length of another string, μ' = 6.8 x 10^-3 kg/m

Let the tension is T'  

Let the length is L.

the formula for the frequency is

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

So, the frequency remains same, length remains same but the tension and the mass per unit length is different.

$\frac{1}{2L}\sqrt{\frac{T}{\mu }}=\frac{1}{2L}\sqrt{\frac{T'}{\mu' }}$

So, $\frac{T'}{\mu '}=\frac{T}{\mu}$

$\frac{T'}{6.8\times 10^{-3}}=\frac{82.4}{2.33\times 10^{-3}}$

T' = 240.5 N