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

Step-by-step 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=(1)/(2L)\sqrt{(T)/(\mu )}`

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

`(1)/(2L)\sqrt{(T)/(\mu )}=(1)/(2L)\sqrt{(T')/(\mu' )}`

So,
`(T')/(\mu ')=(T)/(\mu)`

`(T')/(6.8* 10^(-3))=(82.4)/(2.33* 10^(-3))`

T' = 240.5 N

Answer 2

240.59 N

Step-by-step explanation:

We are given that

Mass per unit length=m/l=
`\mu=2.33* 10^(-3) kg/m`

Fundamental frequency,f=146.8 Hz

Tension=T=82.4 N

Mass per unit length=
`\mu'=6.8* 10^(-3) kg/m`

We have to find the tension.

Velocity,v=
`\sqrt{(T)/(\mu)}=\sqrt{(82.4)/(2.33* 10^(-3))}=188.1m/s`

New tension,T'=
`v^2\mu'=(188.1)^2* 6.8* 10^(-3)`

`T'=240.59 N`

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