# What is the tension of the string needed for a 440 Hz fundamental frequency?

• toothpaste666
In summary, the violin string has a wavelength of .6 meters and the beat frequency is 3.8 Hertz. When the string is tensioned to 34 Newtons, the violinist hears 5.7 beats per second. The frequency of the string is 440 Hertz when it is un-tensioned. Increasing the tension to 38 Newtons will cause the frequency to decrease to 3.8 Hertz.
toothpaste666

## Homework Statement

[/B]
A violin string which is 30 cm long is tuned using a 440 Hz reference tone.

A) What is the wavelength fundamental mode of the string?

B) When the string has a tension of 34 N a violinist hears 5.7 beats per second. What are the frequencies at which the string might be vibrating?

C) The tension is increased and the violinist hears 3.8 beats per second. What is the original frequency of the string?
D) What should the tension of the string be so that the fundamental frequency is 440 Hz

beatf = f2-f1
f= v/2l
lambda = 2l

## The Attempt at a Solution

A) the fundamental wavelength is twice the length of the string. 2 x 30 cm = 60 cm = .6m
B) the difference between the 440 Hz reference tone and the frequency of the violin is 5.7 so the string is either vibrating at 434.3 or 445.7
C) This is where I am stuck. this is my attempt. the beat frequency is 3.8 and the reference tone is 440 so the frequency is now either 443.8 or 436.2 I completely lost on what to do from here though

(C) Does the frequency increase or decrease if you increase tension?

well the velocity of a wave is v=sqrt(FT/u) where FT is the tension and u is the linear density. also v = lambda * f where lambda is the wavelength and f is the frequencey. therefore
lambda * f = sqrt(FT/u)
f = sqrt(FT/u)/lambda
so if the tension was increased the frequency would increase i think

Right. Now you can go back to the frequencies calculated in (B) and see if they are both possible with the additional knowledge from (C).

ahh it can't be 445.7 in part B because it is higher than both the new frequencies and the frequency should have increased if the tension was increased. so part C is 434.3?

so that would mean for part D)
at 34 N of tension:
f*lambda = sqrt(FT/u)
434.3 * .6 = sqrt(34/u)
260.58 = sqrt(34/u)
67902 = 34/u
u= 34/67902 = 5x10^-4 where u is the linear density

if we want f = 440
FT = u (lambda * f)^2 = (5x10^-4) (.6 * 440)^2 = 34.8

is this correct?

Units are missing.
Apart from that, it looks good.

toothpaste666
oh sorry. 34.8 N thanks for your help!

## What is the "Beat Frequency Problem"?

The beat frequency problem is a phenomenon that occurs when two waves with slightly different frequencies interfere with each other. This results in a periodic variation in the intensity of the resulting wave, known as a beat frequency.

## How is beat frequency calculated?

The beat frequency can be calculated by taking the difference between the frequencies of the two interfering waves. For example, if one wave has a frequency of 100 Hz and the other has a frequency of 110 Hz, the beat frequency would be 10 Hz.

## What causes the beat frequency problem?

The beat frequency problem is caused by the superposition of two waves with slightly different frequencies. When these waves interfere, they create a new wave with a periodic variation in intensity that is equal to the difference between their frequencies.

## What are the applications of beat frequency in science?

Beat frequency is used in a variety of scientific fields, including acoustics, optics, and electronics. It is commonly used in music to tune instruments and in radio communication to avoid interference between different frequencies.

## How is the beat frequency problem solved?

The beat frequency problem can be solved by adjusting the frequency of one of the interfering waves. By changing the frequency, the beat frequency can be eliminated or controlled to achieve a desired effect. This is commonly used in audio equipment to tune out unwanted frequencies.

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