Question

the four strings of a bass guitar are 0.865 m long and are tuned to the notes g (98 hz), d (73.4 hz), a (55 hz), and e (41.2 hz). in one bass guitar, the g and d strings have a linear mass density of 4.8 g/m, and the a and e strings have a linear mass density of 29.8 g/m.what is the total force exerted by the strings on the neck?​

Answer 1

Answer: Total force = 636,554.55N

Explanation: To determine tension of strings, wave speed on a string is necessary. Speed is found by:

v = f.λ

f is frequency

λ is wavelength

For the strings, wavelength equals to:

$\lambda = 2L$

L is the length of the bass guitar string

Then, wave speed:

$v=f.2L$

Tension on a string is

$v=\sqrt{\frac{F_{T}}{\mu} }$

$v^{2}=\frac{{F_{T}} }{\mu}$

$F_{T} = v^{2}\mu$

$F_{T} = (2f\lambda)^{2}\mu$

$F_{T} = 4(f\lambda)^{2}\mu$

μ is linear mass density

For g string:

$F_{T} = 4(98.0.865)^{2}.4.8$

$F_{T}$ = 137970.3N

For d string:

$F_{T} = 4(73.4.0.865)^{2}.4.8$

$F_{T}=$ 77397.25N

For a string:

$F_{T} = 4(55.0.865)^{2}.29.8$

$F_{T}=$ 269795N

For e string:

$F_{T} = 4(41.2.0.865)^{2}.29.8$

$F_{T}=$ 151392N

Total force = 137,970.3 + 77,397.25 + 269,795 + 151,392

Total force = 636,554.55N

Total force exerted on the neck by the strings is 636,554.55N.

Answer 2

This question involves the concepts of the tension force in strings, linear mass density, and frequency.

The total force exerted by strings on the guitar is "359.4 N".

The tension force exerted by each string is given as:

$F_T=v^2\mu$

where,

F_T = tension force = ?

v = speed = (frequency)(wavelength) = fλ

μ = linear mass density

Therefore,

$F_T=f^2\lambda^2\mu$

but for strings in this case:

$\lambda = 2(Length of string) = 2(0.865\ m)=1.73\ m$

Therefore,

$F_T=f^2(1.3\ m)^2\mu$

For string g:

$F_{Tg}=(98\ Hz)^2(1.3\ m)^2(4.8\ x\ 10^{-3}\ kg/m)\\F_{Tg}=77.9\ N$

For string d:

$F_{Td}=(73.4\ Hz)^2(1.3\ m)^2(4.8\ x\ 10^{-3}\ kg/m)\\F_{Td}=43.7\ N$

For string a:

$F_{Ta}=(55\ Hz)^2(1.3\ m)^2(29.8\ x\ 10^{-3}\ kg/m)\\F_{Ta}=152.3\ N$

For string e:

$F_{Te}=(41.2\ Hz)^2(1.3\ m)^2(29.8\ x\ 10^{-3}\ kg/m)\\F_{Te}=85.5\ N$

So, the total force will be the sum of all tension forces:

$F=F_{Tg}+F_{Td}+F_{Ta}+F_{Te}$

F = 77.9 N + 43.7 N + 152.3 N + 85.5 N

F = 359.4 N

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