Standing waves can ruin the acoustics of a concert hall if there is excessive reflection of the sound waves that the performers generate. For example, suppose a performer generates a 2093 Hz tone. Of a large amplitude standing wave is present, it is possible for a listener to move a distance of only 4.1 cm and hear the loudness of the tone change from loud to faint. Account for this observation in terms of standing waves, pointing out why the distance is 4.1 cm.

The answer to the questions is;

In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)

The distance 4.1 cm is equivalent to λ/4

Explanation:

For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound

The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ

Therefore 4.1 cm is λ/4

yo yo yo its papaguy in the house lol

okay, ima .

um it shows links to uh so kinda microsoft website is that were ima go or no? ?