Answer :
To play a 528 Hz note on a violin without adjusting the tension in the string, the violinist must press the string so that its vibrating length is approximately 41.67 cm.
The student is seeking to understand the relationship between the frequency of a note played on a violin and the length of the string that vibrates to produce that note. When the tension in the violin string remains constant, the frequency of the note is inversely proportional to the length of the string.
Given that the string produces a 440 Hz note when it is 50 cm long, we need to calculate the length needed for it to produce a 528 Hz note.
The frequency of a string is related to its length by the equation f₁ / f₂ = L₂/ L₁, where f₁ and f₂ are the frequencies and L₁ and L₂ are the lengths of the string for those frequencies. Plugging in the known values (f₁ = 440 Hz, f2 = 528 Hz, L₁ = 50 cm), we can solve for L₂.
Using the equation, we get (440 Hz) / (528 Hz) = L₂ / (50 cm). Simplifying, we find that L₂ = (440/528) * 50 cm, which results in L₂ being approximately 41.67 cm. Therefore, the violinist must press the string at the point where the remaining length is 41.67 cm to play the 528 Hz note.
Answer:
= 0.517 m
Explanation:
This is a resonance exercise where the ends of the string are fixed, therefore it has a node of them, the fundamental (longest) wavelength created has the form
λ = 2L
wave speed is related to wavelength and frequency
v = λ f
v = 2L f
let's calculate
v = 2 0.50 440
v = 440 m / s
since they indicate that the tension of the string does not change and the linear density of the string is constant, the speed of the wave also remains constant
f =[tex]\frac{v}{2L}[/tex]
let's find the length for the new resonance frequency
L = [tex]\ \frac{v}{2f}[/tex]
let's calculate
L = [tex]\frac{440}{2 \ 528}[/tex]
L = 0.5166 m
