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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Apr 2008 18:50:39 IST
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a source s and detector D are initially at a distance of 355 m. both are moving towards one another with same acceleration a =10 m/s2.
frequency of source is = 1 khz the frequency abserved by detector at time t =2 sec is ?
330 sound in air.
35/32 khz
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" From ashes a fire shall be woken From shadows a light shall spring " ---Tolkien
" Throughout the centuries there were men who took first steps down new roads armed with nothing but their own vision. " --- Ayn Rand
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7 Apr 2008 00:10:17 IST
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are you sure with the ans? i don't think so it is right. if deector is moving in +ve x-axis direction &soruce in -ve x-axis direction then for both VS &Vd =a*t=20 fs/V-Vs =(f)d /V+Vd so fd=fs(330+Vd) / 330-Vs fd=330+20 / 330-20 fd=35/31 khz
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7 Apr 2008 17:48:56 IST
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that exactly what i did.
this ans is what is given in brilliant'd 5th full test paper.
!!!!!!!!!
so someone do try & help me
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" From ashes a fire shall be woken From shadows a light shall spring " ---Tolkien
" Throughout the centuries there were men who took first steps down new roads armed with nothing but their own vision. " --- Ayn Rand
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7 Apr 2008 18:40:23 IST
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d ans may b wrong
both r movin wid same velocity
so it is impossible for d ans 2 b 35/31 khz..........
rathin has got d whole soln right.........
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<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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7 Apr 2008 18:43:10 IST
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its a ques from brilliant papre rite????
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The only place where you will find success before hard work is in the dictionary
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7 Apr 2008 18:43:53 IST
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well @nithy u cant do tht way cause Acc is involoved....
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The only place where you will find success before hard work is in the dictionary
A winner never quits.....a quitter never wins......
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12 Apr 2008 21:18:46 IST
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@ ananth_patri
then how to do it?????? THAT's my Q. ....
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