IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: AN10 on March 21, 2010, 12:26:41 pm
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plz solve question no 5 of first variant paper attached below
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For a wave to a have a minima, there must be destructive interference i.e. a phase difference of Pi radians...
However to detect a minima of ZERO, you must have a) A phase difference of Pi Radians b) Same Amplitude
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k i get it.... can u plz solve da second part too
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anyone ??? ???
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I will do it when I get home in 1 hour
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Distance frim S_2 to M is sqrt(1^2 +0,8^)=1.28m
so path difference =0.28m
when f=1kHz wavelength=v/f=330/1000=0.33
When f=4kHz wavelength=330/4000=0.0825
minumum whenw/2=28 or 3w/2 =28 or 5w/2=28 or 7w/2 =28
w=56 or w=18.7 or w=11.2 or w=8
only two of these are between 8.25 and 33
so only two minima
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ok....thanx a lot!!
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Distance frim S_2 to M is sqrt(1^2 +0,8^)=1.28m
so path difference =0.28m
when f=1kHz wavelength=v/f=330/1000=0.33
When f=4kHz wavelength=330/4000=0.0825
minumum whenw/2=28 or 3w/2 =28 or 5w/2=28 or 7w/2 =28
w=56 or w=18.7 or w=11.2 or w=8
only two of these are between 8.25 and 33
so only two minima
i dont get this part only:
minumum whenw/2=28 or 3w/2 =28 or 5w/2=28 or 7w/2 =28
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w is the wavelenfth and mimima where paths difference is w/2 or 3w/2 etc
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cos that means destructive interference