Question

A guitar string is 90 cm long and has a mass of 3.4g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 500N .

What are the frequencies of the fundamental and first two overtones?

Answer 1

Answer:

f = 293.38 Hz

1 st overtone = 586.76 Hz

2 nd overtone = 880.14 Hz

Explanation:

Mass of the string, m =3.4 g = 0.0034 kg

Length of the string, l = 90 cm = 0.90 m

Distance  from the bridge, L = 62 cm = 0.62 m

Tension in the string, T = 500 N

Fundamental frequency is given by:

$f=\frac{\sqrt{Tl/m}}{2L}\\ \Rightarrow f=\frac{\sqrt{500\times0.90/0.0034}}{2\times0.62}=293.38 Hz$

First overtone = 2 × f = 2 × 293.38 =586.76 Hz

Second Overtone = 3×f = 3 × 293.38 = 880.14 Hz

Answer 2

The frequencies of the fundamental and first two overtone are mathematically given as

x1=586.76 Hz

x2= 880.14 Hz

What are the frequencies of the fundamental and first two overtones?

Question Parameter(s):

A guitar string is 90 cm long and has a mass of 3.4g .

The distance from the bridge to the support post is L=62cm, and the string is under a tension of 500N .

Generally, the equation for the   is mathematically given as

$f=\frac{\sqrt{Tl/m}}{2L}$

Therefore

$\f=\frac{\sqrt{500* 0.90/0.0034}}{2*0.62}$

f=293.38 Hz

In conclusion, First overtone

x1=2 × f = 2 × 293.38

x1=586.76 Hz

x2= 3 × 293.38

x2= 880.14 Hz

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