Question

Standing sound waves are produced in a pipe that is 0.8 m long,m open at one end, and closed at the other. For the fundamental and first two overtones, where along the pipe (measured from the closed end) are the displacement antinodes

Answer 1

Positions of antinodes from closed end , in a closed organ pipe are given by ,

$x=L/(2n-1),3L/(2n-1),5L/(2n-1),.............L.$

where $L=$ length of pipe

For first normal mode of vibration (fundamental tone) , $n=1$ , and one antinode is formed ,

$x_1=L/(2\times 1-1)=L=0.8m$ ,

For second normal mode of vibration (first overtone) , $n=2$ , and two antinodes are formed ,

$x_1^{\prime }=0.8/(2\times 2-1)=0.8/3=0.267m$ ,

$x_2^{\prime }=3\times 0.8/(2\times 2-1)=0.24/3=0.8m$ ,

For third normal mode of vibration (second overtone) , $n=3$ , and three antinodes are formed ,

$x_1^{\prime \prime }=0.8/(2\times 3-1)=0.8/5=0.16m$ ,

$x_2^{\prime \prime }=3\times 0.8/(2\times 3-1)=0.24/5=0.48m$ ,

$x_3^{\prime \prime }=5\times 0.8/(2\times 3-1)=4/5=0.8m$