A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has a length of 0.400 m and a mass of 3.00 g. What is the frequency f_1 of the string's fundamental mode of vibration? What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 9781 Hz?
To find the frequency f₁ of the string's fundamental mode of vibration, we can use the equation for the wave speed on a stretched string:
v = √(T/μ)
where:
v is the wave speed,
T is the tension in the string, and
μ is the linear mass density of the string.
The linear mass density μ is given by:
μ = m/L
where:
m is the mass of the string, and
L is the length of the string.
Let's calculate the linear mass density first:
μ = m/L = (3.00 g) / (0.400 m)
μ = 7.50 g/m
Now we can calculate the wave speed v:
v = √(T/μ) = √(765 N / 7.50 g/m)
To ensure consistent units, we'll convert grams to kilograms:
v = √(765 N / 0.00750 kg/m)
Now, calculate the wave speed v:
v ≈ √(102000 / 0.00750) ≈ √13600000 ≈ 3686 m/s
Finally, we can calculate the frequency f₁ of the string's fundamental mode of vibration using the formula:
f₁ = v / λ
where λ is the wavelength. For the fundamental mode, the wavelength is twice the length of the string:
λ = 2L = 2 * 0.400 m = 0.800 m
Now we can calculate the frequency:
f₁ = 3686 m/s / 0.800 m = 4608 Hz
Therefore, the frequency of the string's fundamental mode of vibration (f₁) is 4608 Hz.
To find the number of the highest harmonic that could be heard by a person who can hear frequencies up to 9781 Hz, we need to find the highest harmonic frequency below 9781 Hz.
The frequency of a harmonic is given by the formula:
fₙ = nf₁
where n is the harmonic number.
To find the highest harmonic number below 9781 Hz, we can divide the maximum frequency by the frequency of the fundamental mode:
n = f / f₁ = 9781 Hz / 4608 Hz ≈ 2.122
The highest whole number harmonic that can be heard is 2. Therefore, the number of the highest harmonic that could be heard by a person capable of hearing frequencies up to 9781 Hz is 2.
To find the frequency of the fundamental mode of vibration (f₁) of the piano wire, you can use the formula for the frequency of a vibrating string:
f₁ = (1 / 2L) * √(T / μ)
Where:
- L is the length of the string
- T is the tension of the string
- μ is the linear density of the string
First, let's calculate the linear density of the piano wire:
μ = m / L
Where:
- m is the mass of the wire
Convert the mass from grams to kilograms:
m = 3.00 g = 0.003 kg
Now, you can calculate the linear density:
μ = 0.003 kg / 0.400 m = 0.0075 kg/m
Next, substitute the values of L, T, and μ into the formula for the frequency:
f₁ = (1 / 2 * 0.400 m) * √(765 N / 0.0075 kg/m)
Solve for f₁:
f₁ ≈ 616 Hz
So, the frequency of the string's fundamental mode of vibration is approximately 616 Hz.
To find the number of the highest harmonic that could be heard by a person who can hear frequencies up to 9781 Hz, we need to find the highest harmonic for which the frequency does not exceed 9781 Hz.
The frequency of each harmonic can be calculated using the formula:
fₙ = nf₁
Where:
- fₙ is the frequency of the nth harmonic
- n is the harmonic number
- f₁ is the frequency of the fundamental mode
Let's calculate the number of the highest harmonic:
n = fₙ / f₁
Since we want to find the highest harmonic that can be heard, we start with the highest value of n and decrement it until the frequency of the harmonic does not exceed 9781 Hz.
Let's use a loop to find the highest harmonic:
highest_harmonic = 0
for n in range(1, 10000):
frequency = n * f₁
if frequency <= 9781:
highest_harmonic = n
else:
break
The loop will stop when the frequency of the harmonic exceeds 9781 Hz.
The highest harmonic that can be heard is:
highest_harmonic ≈ 15
Therefore, the number of the highest harmonic that could be heard by a person who can hear frequencies up to 9781 Hz is approximately 15.
http://hyperphysics.phy-astr.gsu.edu/Hbase/waves/string.html
In Part A of this particular problem you are solving for the Fundamental Frequency of the string. Use the equation:
F(fundamental)= [1/(2 * L)] * sqrt[(tension force/mu)]
L= length of string= .4 meters
mu= mass/length=.003 kg/.4=.0075 kg/m
tension force=765 N
plug in knowns and F= 399.2 Hz
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For part B you must solve for the integer multiple, which the problem calls "n". The formula is as follows:
F1= n * F(Fundamental Frequency)
F1=9781 Hz
solve for "n" and n= 24.5....round down so that n=24.