question_answer

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]

A)            20, 20/3, 20/5 etc                

B)            10, 5, 2.5 etc

C)            10, 20, 30 etc                        

D)            15, 25, 35 etc

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\[=\]\[20,\,\,\frac{20}{3},\,\,\frac{20}{5}\,,\,\,\frac{20}{7}\,,\,\,\frac{20}{9}\,,\]?..                    \[=\] \[20\]\[m\], \[6.67\]\[m\], \[4m,\]\[2.85\,m,\] \[2.22\]\[m,\]?..