Homework SourseAn open pipe 095 m long vibrates in the second overtone with
An open pipe, 0.95 m long, vibrates in the second overtone with a frequency of 561 Hz. In this situation, the distance from the center of the pipe to the nearest antinode, in cm is closest to:
I know the answer is 16 but am totaly lost on how to get there.
I know the answer is 16 but am totaly lost on how to get there.
Solution
Wavelength = speed of wave / frequency
= 343 m/s / 561 Hz
= 0.6114 m
= 61.14 cm
At the centre node exist and the distance between node and nearest antinode is always /4
distance d = /4
= 61.14/ 4
= 15.28 cm
˜ 16 cm
