A stretched wire vibrates in its fundamental mode at a frequency of 420 Hz. What would be the fundamental frequency if the wire were one fourth as long, its diameter were quadrupled, and its tension were increased two-fold?
With 4 times the diameter (and assuming the same material density), the mass per unit length would increase by a factor of 16. The ratio (tension)/(linear density) would be 2/16 = 1/8 of its previous value. The wave speed is proportional to the square root of that, or 0.3536 as large as before.
The fundamental frequency is proportional to (wave speed)/(length). This changes by a factor 0.3536/0.250 = 1.414
The new fundamential frequency is 594 Hz.
7,4
To find the fundamental frequency, we need to consider the formula for the frequency of a vibrating stretched wire:
f = (1/2L) * sqrt(T/μ)
Where:
f is the frequency,
L is the length of the wire,
T is the tension in the wire,
and μ is the linear mass density of the wire.
Given:
Frequency (f) = 420 Hz
To find the new fundamental frequency when the wire is one fourth as long, we can substitute L/4 for L:
f' = (1/2(L/4)) * sqrt(T/μ)
Next, if the diameter is quadrupled, it means the radius is doubled. The linear mass density (μ) of the wire is inversely proportional to the square of the radius (r):
μ' = (1/r'^2) * μ
Where:
μ' is the new linear mass density,
r' is the new radius,
and μ is the original linear mass density.
Furthermore, if the tension is increased two-fold, we can substitute 2T for T:
f'' = (1/2(L/4)) * sqrt((2T)/μ')
Now let's simplify the expression:
f'' = (1/2(L/4)) * sqrt((2T)/((1/r'^2) * μ))
f'' = (1/2(L/4)) * sqrt((2T) * (r'^2/μ))
f'' = (1/2(L/4)) * sqrt((2T * r'^2) / μ)
Finally, substitute the given values:
f'' = (1/2(L/4)) * sqrt((2 * 2T * (2r)^2) / μ)
f'' = (1/2(L/4)) * sqrt((8T * 4r^2) / μ)
f'' = (1/2(L/4)) * sqrt((32 * T * r^2) / μ)
Simplifying further, we can simplify the fractions as:
f'' = (1/2L) * (1/2) * sqrt(32 * T * r^2 / μ)
f'' = (1/4L) * sqrt(32 * T * r^2 / μ)
Therefore, the new fundamental frequency (f'') is 1/4L the original frequency (f) and is given by:
f'' = (1/4L) * 420
f'' = 105 / L
So, the new fundamental frequency is 105/L Hz.
To find the fundamental frequency of the wire given these changes, we need to consider the variables that affect the frequency of a stretched wire.
In general, the frequency of vibration of a stretched wire is determined by three variables: the length of the wire (L), the tension in the wire (T), and the mass per unit length (μ or linear density) of the wire.
Given the changes mentioned in the question:
1. The length of the wire is reduced to one-fourth, which means the new length (L') is L/4.
2. The diameter of the wire is quadrupled, which means the new radius (R') is 4 times the original radius (R). Since the diameter is twice the radius, we can say that the new diameter (D') is 8 times the original diameter (D).
3. The tension in the wire is increased two-fold, which means the new tension (T') is 2 times the original tension (T).
To find the new fundamental frequency (f'), we need to consider the relationship between the frequency, length, tension, and mass per unit length of the wire:
f ∝ (1 / L) √(T / μ)
Since the wire is vibrating in its fundamental mode, we can assume that the mass per unit length (μ) is constant and does not change.
Now, let's calculate the changes in the frequency:
1. The length of the wire is reduced to one-fourth, so the new frequency component for length (fL') becomes: fL' = 1 / (L' / L) = 1 / (L/4 / L) = 4.
2. The diameter is quadrupled, so the new frequency component for diameter (fD') becomes: fD' = D' / D = (8R) / R = 8.
3. The tension is increased two-fold, so the new frequency component for tension (fT') becomes: fT' = √(T' / T) = √(2T / T) = √2.
Now, to find the new fundamental frequency (f'):
f' = fL' * fD' * fT' = 4 * 8 * √2 = 32 * √2 ≈ 45.25 Hz.
Therefore, the new fundamental frequency of the wire would be approximately 45.25 Hz.