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?
To find out how much longer the lower-pitched string is, we need to calculate the difference in their lengths.
To do that, we can use the formula for the fundamental frequency of a string:
f = (1/2L) * sqrt(T/μ)
Where:
- f is the fundamental frequency
- L is the length of the string
- T is the tension in the string
- μ is the linear mass density of the string
Since the string tension and the linear mass density are identical for both strings, we can set up the following equation:
(1/2L₁) * sqrt(T/μ) = (1/2L₂) * sqrt(T/μ)
Simplifying the equation, we get:
1/L₁ = 1/L₂
L₂ = L₁ + ΔL
Where L₁ and L₂ are the lengths of the higher-pitched and lower-pitched strings, respectively, and ΔL is the difference in their lengths.
Now we can use the formula for the beat frequency:
f₁ - f₂ = Δf
Where:
- f₁ is the frequency of the higher-pitched string
- f₂ is the frequency of the lower-pitched string
- Δf is the beat frequency
Given that f₁ = 400 Hz and Δf = 4 Hz, we can rearrange the formula to solve for f₂:
f₂ = f₁ - Δf = 400 Hz - 4 Hz = 396 Hz
Plugging this value into the previous equation, we have:
1/L₁ = 1/L₂
1/L₁ = 1/(L₁ + ΔL)
To solve for ΔL, we can set up the following equation:
1/(L₁ + ΔL) - 1/L₁ = 0
Multiplying through by L₁(L₁ + ΔL), we get:
L₁ - (L₁ + ΔL) = 0
Simplifying, we find:
-ΔL = -L₁
Dividing by -1 on both sides, we have:
ΔL = L₁
Now, we can substitute the value of L₁ into the equation:
ΔL = L₁ = (1/f₂) * v
Where:
- v is the speed of transverse waves in the wire
Given that f₂ = 396 Hz and v = 200 m/s, we can calculate ΔL:
ΔL = (1/396 Hz) * 200 m/s
ΔL ≈ 0.505 m
Therefore, the lower-pitched string is approximately 0.505 meters longer than the higher-pitched string.
To answer this question, we need to use the formula for the beat frequency of two waves. But before we can do that, we need to understand what beat frequency means.
Beat frequency is the difference in frequency between two waves. When two waves of slightly different frequencies interfere with each other, they produce a fluctuation in the overall sound intensity, which is what we perceive as beats. The beat frequency is equal to the difference in frequency between the two waves.
Now let's use the formula for beat frequency:
Beat frequency (fb) = |f1 - f2|,
where f1 and f2 are the frequencies of the two waves.
In this case, the beat frequency is given as 4Hz, and one of the frequencies (f1) is 400Hz. We can rearrange the formula to solve for the unknown frequency (f2):
f2 = f1 ± fb.
In this formula, the ± symbol indicates that f2 can be either slightly higher or slightly lower than f1, depending on whether the waves are interfering constructively or destructively to produce the beats. Since we are looking for the lower-pitched string, we need to subtract the beat frequency (4Hz) from the higher frequency (400Hz):
f2 = 400Hz - 4Hz = 396Hz.
Now that we have the frequency of the lower-pitched string, we can use the formula for the speed of a wave on a string to find the length ratio between the two strings.
The formula for the speed of a wave (v) on a string is:
v = √(T/μ),
where T is the tension in the string and μ is the linear mass density of the string.
Since both strings are identical, they have the same tension (T) and linear mass density (μ). Therefore, the speed (v) of the waves on both strings is the same (200 m/s).
Now, let's use the formula for wave speed to find the length ratio:
v = √(T/μ).
Since the tension and linear mass density are the same for both strings, we can rewrite the formula as:
v = √(T1/μ) = √(T2/μ),
where T1 and T2 are the tensions in the higher-pitched and lower-pitched strings, respectively.
We can square both sides of the equation to get rid of the square root:
v^2 = T1/μ = T2/μ.
Since v is the same for both strings, we can cancel it out:
T1/μ = T2/μ.
Cancelling out μ, we get:
T1 = T2.
This means that the tensions in both strings are the same.
The length of a string is directly proportional to its fundamental frequency. So, if the frequency of the lower-pitched string is 396Hz (slightly lower than 400Hz), we can calculate the corresponding length (L2) using the frequency-length relationship:
L2 = (v/f2),
where v is the speed of transverse waves on the string (200 m/s).
L2 = (200 m/s)/(396Hz) = 0.505 m.
Therefore, the length of the lower-pitched string is 0.505 meters longer than the higher-pitched string.