A guitar string is struck and found have a frequency of 2048Hz. If both tension and length are doubled, what is the new frequency of the string?
The fundamental frequency (first harmonic) of a plucked string is given by
f=(1/2L)√(T/ρ)
where L=length, T= tension and ρ=mass/unit length
So we see that the frequency is inversely proportional to the length, and proportion to the square-root of the tension.
If both of them are doubled, then new frequency, f1 is can be obtained by :
f1=f*(1/2)*&radic(2)
=f/&radic(2)
=2048/1.414213562
=1448 Hz approximately
To find the new frequency of a guitar string when both tension and length are doubled, we need to understand the relationship between frequency, tension, and length of a string.
The frequency of a vibrating string is determined by the following formula:
f = (1 / 2L) * sqrt(T / μ)
Where:
f = frequency of the string
L = length of the string
T = tension in the string
μ = linear density of the string
Given that the tension and length of the string are doubled, we can express the new tension as 2T and the new length as 2L.
Substituting these values into the formula, we can solve for the new frequency:
f' = (1 / 2(2L)) * sqrt(2T / μ)
Simplifying further:
f' = (1 / 4L) * sqrt(2T / μ)
f' = 1 / 4 * (1 / L) * sqrt(2T / μ)
Since we know the initial frequency (f = 2048Hz), we can substitute it into the equation to solve for μ:
2048 = 1 / 4 * (1 / L) * sqrt(2T / μ)
Now, let's assume that the linear density (μ) remains the same when the tension and length are doubled. In reality, the linear density of a string can change slightly when tension or length is changed, but for the purpose of this question, let's assume it remains constant.
So we can take μ out of the equation:
2048 = 1 / 4 * sqrt(2T / μ) * (1 / L)
We can rearrange the equation to solve for f':
f' = 2048 * 4 * L / sqrt(2T)
Now, we can plug in the values given in the question: f = 2048Hz, L = 2L, and T = 2T:
f' = 2048 * 4 * (2L) / sqrt(2 * 2T)
Simplifying further:
f' = 8192 * L / sqrt(4T)
Since both the tension and length are doubled, we can simplify further:
f' = 8192 * L / (2 * sqrt(T))
Finally, we can simplify the equation to find the new frequency:
f' = 4096 * L / sqrt(T)
Therefore, when both tension and length are doubled, the new frequency of the string is 4096Hz.
Well, if the tension and length are doubled, it sounds like the poor string is having a bit of an identity crisis. It's getting stretched all over the place! But fear not, dear string, for I have the answer to your question.
When tension and length are both doubled, the frequency of the string is inversely proportional to the square root of the product of the tension and the length. So, let's do some calculations while offering emotional support to the string.
If the initial frequency is 2048Hz, and both the tension and the length are doubled, it means we have:
New frequency = Initial frequency * √((2 * new tension) / (2 * new length))
Now, let's simplify this expression and make the calculations:
New frequency = 2048Hz * √(2/2)
Drumroll, please...
The new frequency of the string is still 2048Hz! So, even though the tension and length have changed, the frequency remains the same. It looks like that string has some serious stability!
Keep on strumming, my friend!
To determine the new frequency of the guitar string after both tension and length are doubled, we can use the equation for the frequency of a vibrating string:
f = (1/2L) * sqrt(Tension/μ),
where:
f = frequency,
L = length of the string,
Tension = tension on the string, and
μ = linear mass density of the string.
We know the initial frequency (f1) of the string is 2048Hz. Let's assume the initial tension (T1) and length (L1) are T and L respectively.
So, f1 = (1/2L1) * sqrt(T1/μ).
If both tension and length are doubled, the new tension (T2) and length (L2) can be expressed as 2T and 2L respectively.
Therefore, the new frequency (f2) can be calculated as follows:
f2 = (1/2L2) * sqrt(T2/μ)
= (1/2(2L)) * sqrt(2T/μ)
= (1/4L) * sqrt(2T/μ)
= (1/4) * (1/2L1) * sqrt(2T/μ)
= (1/8L1) * sqrt(2T/μ).
Since f1 = 2048Hz, we can substitute this value into the equation to solve for f2:
2048 = (1/8L1) * sqrt(2T/μ).
Now, since both T and L double, and assuming the linear mass density (μ) remains constant, we can substitute 2T for T and 2L1 for L1:
2048 = (1/8(2L1)) * sqrt(2(2T)/μ)
= (1/16L1) * sqrt(4T/μ).
Simplifying further, we have:
2048 = (1/16L1) * 2 * sqrt(T/μ)
= (1/8L1) * sqrt(T/μ).
Finally, multiplying both sides of the equation by 8L1 and squaring both sides, we get:
16384L1 = T/μ.
Therefore, the new frequency (f2) of the string after doubling both tension and length is the same as the initial frequency, which is 2048Hz.