question_answer

A man is watching two trains, one leaving and the other coming in with equal speeds  of 4 m/sec. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air = 320 m/sec) will be equal to                                       [NCERT 1984; CPMT 1997; MP PET 1999; RPMT 2000; BHU 2004, 05]

A)            6    

B)            3

C)            0    

D)            12

Correct Answer: A

Solution :

                                      Frequency of sound heard by the man from approaching train                    \[{{n}_{a}}=n\,\left( \frac{v}{v-{{v}_{s}}} \right)=240\,\left( \frac{320}{320-4} \right)=243\,Hz\]                    Frequency of sound heard by the man from receding train  \[{{n}_{r}}=n\,\left( \frac{v}{v+{{v}_{s}}} \right)=240\,\left( \frac{320}{320+4} \right)=237Hz\]                    Hence, number of beats heard by man per sec                    \[={{n}_{a}}-{{n}_{r}}=243-237=6\]                    Short trick : Number of beats heard per sec                    \[=\frac{2nv{{v}_{S}}}{{{v}^{2}}-v_{S}^{2}}=\frac{2nv{{v}_{S}}}{(v-{{v}_{S}})(v+{{v}_{S}})}=\frac{2\times 240\times 320\times 4}{(320-4)(320+4)}=6\]