00:01
Hello students, in this question there are two loudspeaker which are separated by the distance d is equal to 3 meter.
00:10
A listener is listening from the speaker 1 at 4 meter and realizing that the sound intensity is minimum at 4 meter and he is crossing over the distance in arc shape and maintaining the same distance of 4 meter from the speaker 1 so that the distance between the listener and the speaker 2 alone will be increasing.
00:36
So now we have to find out at which distance the listener is from the speaker s2 when the first maximum sound intensity occurs.
00:49
So we are given that the velocity of the sound is 340 meter per second.
00:56
So the listener is realizing that the speaker gives its minimum intensity when the distance is 4.
01:03
Therefore, at minimum we have del d minimum is equal to 2k plus 1 into lambda by 2 and for the next maximum del d max is equal to this del d minimum plus lambda by 2.
01:30
So for this from the pythagoras theorem we can calculate that s2 into l is equal to so here root of 3 squared plus 4 squared.
01:50
So 3 squared is 9, 4 squared is 16 so which is equal to root 25 therefore s2 l is equal to 5 meter.
02:05
So we have to calculate the path difference when the first minimum sound intensity is heard.
02:12
So path difference when therefore, the path difference can be calculated as s2 l minus s1 l which is equal to lambda by 2 into 2k plus 1.
02:35
This is 5 meter minus 4 meter which is equal to lambda by 2 where k is equal to 0 and this is 1 meter is equal to lambda by 2 this implies lambda is equal to 2m the path difference is 2 meter.
02:56
The next maximum now the next maximum is obtained such that del d max is equal to del d minimum plus lambda by 2...