Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for DAC:
⇒
Now for the BAC:
⇒
Now, differentiating w.r.t x:
For maximum angle, = 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 =
After solving the above eqn, we get
x =
The observer should stand at a distance equal to x =
Answer:6.298,4.1,1.9015
Explanation:
Wavelength=
==4.397 m
Distance of 3rd speaker from speaker A is x
From B 78-x
Difference between the distances must be a whole number of wavelengths
First
for 1 st wavelength
2x=8.2+4.397=12.597
x=6.298m
second
For zero wavelength
2x=8.2
x=4.1m
Third
x=1.9015 m
Answer:
a). 100.2 MHz (typical frequency for FM radio broadcasting)
The wavelength of a frequency of 100.2 Mhz is 2.99m.
b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)
The wavelength of a frequency of 1070 khz is 280.3 m.
c. 835.6 MHz (common frequency used for cell phone communication)
The wavelength of a frequency of 835.6 Mhz is 0.35m.
Explanation:
The wavelength can be determined by the following equation:
(1)
Where c is the speed of light, is the wavelength and
is the frequency.
Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.
<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>
Then, can be isolated from equation 1:
(2)
since the value of c is . It is necessary to express the frequency in units of hertz.
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.
<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>
<em> </em>
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 1070 khz is 280.3 m.
<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.