wave_mechanics critical_thinking Question 4
Question: In a large room, a person receives direct sound waves from a source 120 metres away from him. He also receives waves from the same source which reach him, being reflected from the 25 metre high ceiling at a point halfway between them. The two waves interfere constructively for wavelength of
[Roorkee 1982]
Options:
A) 20, 20/3, 20/5 etc
B) 10, 5, 2.5 etc
C) 10, 20, 30 etc
D) 15, 25, 35 etc
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ S $ be source of sound and $ P $ the person or listner.
The waves from $ S $ reach point $ P $ directly following the path $ SMP $ and being reflected from the ceiling at point $ A $ following the path $ SAP $ .
$ M $ is mid-point of $ SP $ (i.e.
$ SM=MP) $ and $ \angle SMA=90^{o} $ Path difference between waves $ \Delta x=SAP-SMP $ We have $ SAP=SA+AP=2(SA) $
$ = $
$ 2\sqrt{[{{(SM)}^{2}}+{{(MA)}^{2}}]} $ = $ 2\sqrt{(60^{2}+25^{2})} $ =130 $ m $
$ \therefore $ Path difference = SAP ? SMP $ =130-120=10 $ m Path difference due to reflection from ceiling = $ \frac{\lambda }{2} $
$ \therefore $ Effective path difference Dx = $ 10+\frac{\lambda }{2} $ For constructive interference $ \Delta x=10+\frac{\lambda }{2}=n\lambda \Rightarrow (2n-1)\frac{\lambda }{2}=10(n=1,2,3….) $
$ \therefore $ Wavelength $ \lambda =\frac{2\times 10}{(2n-1)}=\frac{20}{2n-1} $ .
The possible wavelength are l$ = $
$ 20,\frac{20}{3},\frac{20}{5},\frac{20}{7},\frac{20}{9}, $ ?..
$ = $
$ 20 $
$ m $ , $ 6.67 $
$ m $ , $ 4m, $
$ 2.85m, $
$ 2.22 $
$ m, $ ?..
BETA
