a pipe that is 0.46 m long and open at both ends vibrates in the second overtone with a frequency of 1150 hz. in this situation, the distance from the center of the pipe to the nearest antinode is closest to group of answer choices zero. 7.7 cm. 15 cm. 12 cm. 3.8 cm.

Question

21-10-2022
Physics

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a pipe that is 0.46 m long and open at both ends vibrates in the second overtone with a frequency of 1150 hz. in this situation, the distance from the center of the pipe to the nearest antinode is closest to group of answer choices zero. 7.7 cm. 15 cm. 12 cm. 3.8 cm.

Answer :

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aditmarkas98 aditmarkas98 · Answer 1 · 2022-10-25T07:06:59-04:00

The distance of the center pipe to the nearest antinode at the second harmonic frequency is closest to 12 cm.

We need to know about the harmonic frequency to solve this problem. When the string is vibrating at its second harmonic frequency, it should follow the rule

λ = L

where λ is wavelength and L is the length of the string

From the question above, we know that

f = 1150 Hz

L = 0.46 cm

Find the wavelength

λ = L

λ = 0.46 m

The distance from the center of the pipe to the nearest antidote is equal to 1/4 of the wavelength. Hence,

s = 1/4 λ

s = 1/4 . 0.46

s = 0.115 m

s ~ 12 cm

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