An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat frequency is generated by the first overtone of the closed pipe with the fundamental of the open pipe

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

        (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

Closed tube:

L': length of the closed tube = 0.702821m

Next, you use the following formula for the beat frequency:

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer