Apipe 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?

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 L’ be the length that was cut from the first pipe.

So, the length of the second pipe is L – L’

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

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

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm  

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

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s