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A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shortened by cutting off a portion of the open end. Now when both pipes vibrate at their fundamental frequencies, a beat frequency of 10 Hz is heard. How many centimeters were cut off the end of the second pipe?

Respuesta :

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

Since the speed of sound is not given in the question, we would assume it to be 340m/s

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider Lā€™ be the length that was cut from the first pipe.

So, the length of the second pipe is L ā€“ Lā€™

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - Lā€™)

The beat frequency due to the fundamental frequencies of the first and second pipe is

f2 ā€“ f1 = 10hz

[v/4(L - Lā€™)] ā€“ 266 = 10

[v/4(L ā€“ Lā€™)] = 10 + 266

[v/4(L ā€“ Lā€™)] = 276

(L - Lā€™) = v/(4 x 276)

(L ā€“ Lā€™) = 340/(4 x 276)

(L ā€“ Lā€™) = 0.30797

Lā€™ = 0.31954 ā€“ 0.30797

Lā€™ = 0.01157 m = 1.157 cm ā‰… 1.16cm Ā 

Hence, 1.16 cm were cut from the end of the second pipe

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