Consider two point sources separated by a distance of 10 met

Consider two point sources separated by a distance of 10 meters. The longitudinal waves emitted from the two sources have the wavelength of = lm. The velocity of the wave is 2m/sec. What is the frequency f and the period T?

Solution

Frequency=Velocity/wavelength

=(2m/sec)/1m

=2 Hz

Time period= 1/Frequency

=1/2

=0.5 sec

a)

Amplitude at C =0.1+0.1=0.2 m (Since waves are in phase)

b)

Waveform will be at same amplitude that is 0.2 m

c)

If the distance between the points A and B is changed. The waves interfere distructively. Therefore, amplitude is zero. That is 0.1-0.1=0

d)

If source A is delayed by 0.5 sec

The path difference is= 2(0.5)=1 m

Therefore, waves interfere constructevely.

The wave amplitude at point C is.,

0.1+0.1=0.2 m