stationary wave help

Two loudspeakers face each other and are seperated by a distance of about 20m. They are connected to the same oscillator, which gives a signal frequency of 800Hz.
a) calculate the seperation of adjacent nodes along the line joining the two loudspeakers.
b) A small microphone, moved at constant speed along this line records a signal which varies periodically at 5.0Hz. Calculate the speed at which the microphone moves. Assume that the speed of sound is 350ms.

any help would be appreciated!

Reply 1

Morbo

a) Node separation is (lambda/2) = c/2f = (350ms^-1 / 2*800Hz) = 0.21875m = 0.22m

b) To go from zero signal to max signal and back to zero again is the distance of the node separation (lambda/2). It travels this distance 5 times a second (i.e. at 5.0Hz). Hence:

v = f*(lambda/2) = 5.0Hz * 0.21875m = 1.09375 ms^-1 = 1.1 ms^-1

Reply 2

cat123