P18_080 - f = 482 Hz. (b) In this case, neither tension nor

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80. (a) We proceed by dividing the (velocity) equation involving the new (fundamental) frequency f 0 by the equation when the frequency f is 440 Hz to obtain f 0 λ = v u u t τ 0 µ τ µ = f 0 f = r τ 0 τ where we are making an assumption that the mass-per-unit-length of the string does not change signiFcantly. Thus, with τ 0 =1 . 2 τ ,wehave f 0 / 440 = 1 . 2. Therefore,
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Unformatted text preview: f = 482 Hz. (b) In this case, neither tension nor mass-per-unit-length change, so the wavespeed v is unchanged. Hence, f = f = f (2 L ) = f (2 L ) where Eq. 18-38 with n = 1 has been used. Since L = 2 3 L , we obtain f = 3 2 (440) = 660 Hz....
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