Question

Two sound wave of frequency 100 hz and 102 hz and having same amplitude a are interfering. A detector which can detect waves of amplitude grater than or equal to a is kept at rest. In a time interval of 12 seconds find the total duration in which detector is active

Answer 1

In 12 seconds interval, active time of detector = 8 seconds.

Explanation:

y1 = ASinω1t

y2 = ASinω2t

y1 = 2A.Cos (ω2 - ω1 /2). Sin(ω2 + ω1 /2)

So the resultant amplitude Ar will be:

Ar = 2Ao.I.Cos (△ω)t/2I

(△ω)t/2 = π/2 implies t = 1/4 seconds

(△ω)t/2 = π/3  implies t = 1/6 seconds

In one cycle of intensity of 1/2 seconds, the detector remains idle for:

2 (1/4 - 1/2) = 1/6 seconds

Therefore, In 1/2 second cycle, active time will be 1/2 - 1/6 = 1/3 seconds

In 12 seconds interval, active time will be 12 * (1/3 / 1/2) = 8 seconds.