An industrial firm has two electrical loads connected in parallel across the power source. Power is supplied to the firm at 4000
1 answer:
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Answer:
The minimum frequency is 702.22 Hz
Explanation:
The two speakers are adjusted as attached in the figure. From the given data we know that
=3m
=4m
By Pythagoras theorem
Now
The intensity at O when both speakers are on is given by
Here
- I is the intensity at O when both speakers are on which is given as 6
- I1 is the intensity of one speaker on which is 6
- δ is the Path difference which is given as
- λ is wavelength which is given as
Here
v is the speed of sound which is 320 m/s.
f is the frequency of the sound which is to be calculated.
where k=0,1,2
for minimum frequency , k=1
So the minimum frequency is 702.22 Hz
The frequency of the wave is
Explanation:
The frequency, the wavelength and the speed of a wave are related by the following equation:
where
c is the speed of the wave
f is the frequency
is the wavelength
For the radio wave in this problem,
Therefore, the frequency is:
Learn more about waves here:
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