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Physics
There are three sources of sound of equal intensity with frequencies 400,401 and 402vibsec- 1. The number of beats heard per second is:
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Q. There are three sources of sound of equal intensity with frequencies $400,401$ and $402vibsec^{- 1}.$ The number of beats heard per second is:
NTA Abhyas
NTA Abhyas 2020
A
$0$
B
$1$
C
$2$
D
$3$
Solution:
For source of frequency, $400Hz$ , $y_{1}=asin2\pi \left(f - 1\right)t$
For source of frequency, $401Hz$ , $y_{2}=asin2\pi ft$
For source of frequency, $402Hz$ , $y_{3}=asin2\pi \left(f + 1\right)t$
Resultant of these waves,
$Y=y_{1}+y_{2}+y_{3}\Rightarrow Y=asin2\pi \left(f\right)t+\left[asin 2 \pi \left(f - 1\right) t + asin 2 \pi \left(f + 1\right) t\right]\Rightarrow Y=a\left(1 + cos 2 \pi t\right)sin2\pi ft$
For maximum amplitude,
$cos2\pi t=1\Rightarrow 2\pi t=2\pi m\Rightarrow t=m\left[m = 0 , 1 , 2 , 3 . . . .\right]$
For minimum amplitude,
$cos2\pi t=-\frac{1}{2}\Rightarrow 2\pi t=2\pi m+\frac{2 \pi }{3}\Rightarrow t=m+\frac{1}{3}\left[m = 0 , 1 , 2 , 3 . . . .\right]t=\frac{1}{3},\frac{4}{3},\frac{7}{3},.......$
Time between successive minima, $t_{2}-t_{1}=\frac{4}{3}-\frac{1}{3}=1$
Hence, number of beats per second is $1$ .