A pipe of length 2.7 m is closed at one end and sustains a standing wave at its second overtone. Determine the distance between a node and an adjacent antinode.

The distance between an antinode and adjacent node is 0.54m

Explanation:

Given:

Length of the pipe, L = 2.7m

Pipe is closed at one end

Standing wave is at 2nd overtone

Distance between a node and adjacent antinode, d = ?

In a pipe, closed at one end and open at one end, the 1st overtone is the 3rd harmonic( n=3 ) and 2nd overtone is the 5th harmonic(n=5).

Antinodes and nodes are always formed equidistant.

So, the distance between an antinode and node would be one-fifth of the length of the pipe.

So, the distance would be    ( n = 5)

Thus, the distance between an antinode and adjacent node is 0.54m

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//