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horsena [70]
3 years ago
15

What is the lowest frequency that will resonate in an organ pipe 2.00 m in length, closed at one end? The speed of sound in air

is 340 m/s.
Physics
1 answer:
Paraphin [41]3 years ago
7 0

Answer:

42.5 Hz.

Explanation:

The fundamental frequency of a closed pipe is given as

f₀ = v/4l....................... Equation 1

Where f₀ = lowest frequency, v = speed of sound in air, l = length of the organ pipe

Given: v = 340 m/s, l = 2.00 m.

Substitute into equation 1

f₀ = 340/(4×2)

f₀ = 340/8

f₀ = 42.5 Hz.

Hence the smallest frequency that will resonant in the organ pipe = 42.5 Hz.

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<h3>How to solve for the time interval</h3>

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Read more on waves here

brainly.com/question/25699025

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complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

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