Two identical guitar strings are stretched with the same tension between supports that are not the same distance apart. The fundamental frequency of the higher-pitched string is 400Hz, and the speed of transverse waves in both wires is 200 m/s. How much longer is the lower-pitched string if the beat frequency is 4Hz?
see other post on this.
do you have a answer or formula to this bobpursley?
i went to other post on this an it redirected me to this one...if you don't mind i really would appreciate if you re-answer the question
To find the length difference between the two strings, we need to use the formula for beat frequency:
Beat frequency (f_beat) = |f1 - f2|
Where f1 and f2 are the frequencies of the two strings.
In this case, we have:
f1 = 400 Hz (higher-pitched string)
f_beat = 4 Hz
To find the frequency of the lower-pitched string (f2), we can use the formula:
f = v / λ
Where f is the frequency, v is the speed of transverse waves, and λ is the wavelength.
Since the strings are identical and have the same tension, the speed of transverse waves (v) is the same for both strings, which is 200 m/s.
Now, let's calculate the wavelength of the higher-pitched string (λ1):
λ1 = v / f1
= 200 m/s / 400 Hz
= 0.5 m
Since the two strings have the same tension, the speed of transverse waves (v) and the density of the strings must be the same. Therefore, their wavelengths (λ) are inversely proportional to their frequencies.
Now, let's calculate the wavelength of the lower-pitched string (λ2):
λ2 = λ1 * f1 / f2
Substituting the given values, we have:
0.5 m = (0.5 m) * 400 Hz / f2
Simplifying the equation:
f2 = 800 Hz
Now, we can find the length (L2) of the lower-pitched string using the formula:
L2 = λ2 * (n + 0.5)
Where L2 is the length of the lower-pitched string, λ2 is the wavelength of the lower-pitched string, and n is the number of antinodes (0.5 represents the position of the first antinode).
Since we are dealing with the fundamental frequency, there is only one antinode, so n = 0.
L2 = λ2 * (0 + 0.5)
= λ2 * 0.5
= 0.25 m
Therefore, the longer-pitched string is 0.25 meters longer than the lower-pitched string if the beat frequency is 4 Hz.