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A standing wave is set up in a 2.0 m length string fixed at both ends. The string is then made to vibrate in 5 distinct segments when driven by a 120 Hz source. What is the natural, fundamental frequency of this string?

Answer :

Answer:

The natural, fundamental frequency of this string is 24 hertz.          

Explanation:

Length of the string, l = 2 m

Number of segments, n = 5

Frequency, f = 120 Hz

Let f' is the natural fundamental frequency of this string. The frequency for both side ended string is given by :

[tex]f=\dfrac{nv}{2l}[/tex]

[tex]v=\dfrac{2fl}{n}[/tex]

[tex]v=\dfrac{2\times 120\times 2}{5}[/tex]        

v = 96 m/s

For fundamental frequency, n = 1

[tex]f'=\dfrac{v}{2l}[/tex]

[tex]f'=\dfrac{96}{2\times 2}[/tex]

f' = 24 Hertz

So, the natural, fundamental frequency of this string is 24 hertz. Hence, this is the required solution.

Lanuel

The natural, fundamental frequency of this standing wave is equal to 24 Hertz.

Given the following data:

Length of the string = 2.0 meters.

Number of segments = 5.

Frequency = 120 Hertz.

How to calculate natural, fundamental frequency.

First of all, we would determine the velocity of the standing wave by using this formula:

[tex]F=\frac{nV}{2L} \\\\V=\frac{2FL}{n} \\\\V=\frac{2 \times 120 \times 2.0}{5}\\\\V=\frac{480}{5}[/tex]

V = 96 m/s.

Now, we can calculate the natural, fundamental frequency by using this formula:

[tex]F'=\frac{V}{2L} \\\\F'=\frac{96}{2 \times 2.0} \\\\F'=\frac{96}{4.0}[/tex]

F' = 24 Hertz.

Read more on frequency here: brainly.com/question/3841958

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