in the resultant intensity due to the superposition of two waves of amplitude 3 and 4 in opposite phase is 1 and wave of amplitude 4 is stopped, the intensity at that point will be.
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0
Answer:
Given :
y
1
=5sin(wt+kx)
y
2
=5cos(wt+kx+150
∘
)
y=y
1
+y
2
where y is the resultant wave
The amplitude of this resultant wave is given by the formula:
A=
(A
1
)
2
+(A
2
)
2
+2A
1
A
2
cosϕ
Here;
A
1
=5
A
2
=5
ϕ=150
∘
on substituting the values on the given formula:
A=
(5)
2
+(5)
2
+2×5×5×cos150
∘
=
25+25−2×25×
2
1
..(cos150
∘
=−
2
1
)
Hence A=5 (option A is correct)
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