There are three sources of sound of equal intensity with frequencies $$400$$, $$401$$ and $$402$$ $$vib/$$$$sec$$. The number of beats heard per second is

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Infinity Learn · Accepted Answer

Let n – $$1$$ $$(=$$ $$400)$$, n $$(=$$ $$401)$$ and n $$+$$ $$1$$ $$(=$$ $$402)$$ be the frequencies of the three waves. If a be the amplitude of each then $$y=asin2$$π(n$$−$$1)t$$, $$y=asin2$$$$π$$nt and $$y3=asin2$$π$$(n+1)tResultant$$ displacement due to all three waves is $$y=y1+y2+y3=asin2$$$$π$$$$nt+a$$ [$$sin2$$π$$(n$$−$$1)t+sin2$$π$$(n+1)t$$]$$=asin2$$$$π$$$$nt+a$$ [$$2sin2$$$$π$$$$ntcos2$$$$π$$t]Using $$sinC+sinD=2sinC+D2cosC$$−$$D2$$⇒$$y=a$$ $$(1+2cos2$$$$π$$$$t)sin2$$$$π$$ntThis is the resultant wave having amplitude $$=(1+2cos2$$$$π$$$$t)For$$ maximum amplitude $$cos$$ $$2$$$$π$$t $$=$$ $$1$$ ⇒ $$2$$$$π$$t $$=$$ $$2m$$π where m $$=$$ $$0$$, $$1$$, $$2$$, $$3$$, ...⇒ t $$=$$ $$0$$, $$1$$, $$2$$, $$3$$ ...Hence time interval between two successive maximum is $$1$$ $$sec$$. So beat frequency $$=$$ $$1Also$$ for minimum amplitude $$(1+2cos$$ $$2$$$$π$$$$t) $$=$$ $$0$$ ⇒$$cos2$$$$π$$$$t=$$−$$12$$  ⇒$$2$$$$π$$$$t=2m$$π$$+2$$$$π$$$$3$$⇒$$t=m+13$$⇒$$t=13$$,$$43$$,$$73$$,$$103$$,....       (for$$ m $$=$$ $$0$$, $$1$$, $$2$$, ..$$) Hence time interval between two successive minima is $$1$$ $$sec$$ so, number of beats per second $$=$$ $$1Note$$ : $$PET/PMT$$ Aspirants can remember result only.

There are three sources of sound of equal intensity with frequencies 400, 401 and 402 vib/sec. The number of beats heard per second is

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