Answer:
25.4 Hz
Explanation:
The fundamental frequency (first harmonic) of a string tied down at both ends is given by
[tex]f=\frac{1}{2L}\sqrt{\frac{T}{m/L}}[/tex]
where
L is the length of the string
T is the tension in the string
m is the mass of the string
The equation can be rewritten as
[tex]fL^{3/2}=\frac{1}{2}\sqrt{\frac{T}{m}}[/tex]
We notice that the term on the right remains constant for this string (we assume the tension remains the same), so we can write
[tex]f_1 L_1^{3/2}=f_2 L_2^{3/2}[/tex]
where in this case:
[tex]f_1=41 Hz[/tex] is the fundamental frequency at the beginning
[tex]L_1=0.64 m[/tex] is the initial length of the string
[tex]L_2=0.88 m[/tex] is the final length of the string
Solving for [tex]f_2[/tex], we find the final fundamental frequency:
[tex]f_2=\frac{f_1 L_1^{3/2}}{L_2^{3/2}}=\frac{(41)(0.64)^{3/2}}{0.88^{3/2}}=25.4 Hz[/tex]
Answer:
The correct answer is 29.8
Explanation:
find velocity of the first length and frequency
v = wavelength * f = .64 * 41
v = 26.24 m/s
then, divide by the new length
26.24 / .88 = 29.81 or 30
