Question
What do you understand by beat? Explain beats analytically.
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Answer
For definition, see text: Consider two simple harmonic progressive waves travelling simultaneously in the same direction and in the same medium. Let 'A' be the amplitude of each wave. There is no initial phase difference between them. v1 and v2 be their frequencies. If y1 and y2 be displacements of the two waves, then $\text{y}_1=\text{A}\sin2\pi\text{v}_1\text{t}$ and $\text{y}_2=\text{A}\sin2\pi\text{v}_2\text{t}$ If y be the result and displacement at any instant, then $\text{y}=\text{y}_1+\text{y}_2=\text{A}\sin(2\pi\text{v}_1\text{t})+\sin(2\pi\text{v}_2\text{t})$ $=\text{A}\Bigg[2\sin\Big(\frac{2\pi(\text{v}_1+\text{v}_2)\text{t}}{2}\Big)\cos\Big(\frac{2\pi(\text{v}_1-\text{v}_2)\text{t}}{2}\Big)\Bigg]$ $=2\text{A}\cos\pi(\text{v}_1\text{v}_2)\text{t}\sin \pi(\text{v}_1+\text{v}_2)\text{t}$ $=\text{R}\sin\pi(\text{v}_1+\text{v}_2)\text{t}\ \dots(1)$ where $\text{R}=2\text{A}\cos\pi(\text{v}_1-\text{v}_2)\text{t}\ \dots(2)$ is the amplitude of the resultant displacement and depends upont. The following cases arise.
- If R is maximum, then
$\cos\pi(\text{v}_1-\text{v}_2)\text{t}=\text{max}.=\pm1=\cos\text{n}\pi$
$\therefore \pi(\text{v}_1-\text{v}_2)\text{t}=\text{n}\pi$
$\text{t}=\frac{\text{n}}{\text{v}_1-\text{v}_2}\ \dots(3)$
where n = 0, 1, 2...
$\therefore$ Amplitude becomes maximum at times given by
$\text{t}=0,\frac{1}{\text{v}_1-\text{v}_2},\frac{2}{\text{v}_1-\text{v}_2}=\frac{3}{\text{v}_1-\text{v}_2},...$
$\therefore$ Time interval between two consecutive maxima is
$=\frac{1}{\text{v}_1-\text{v}_2}$
$\therefore$ Beat period $=\frac{1}{\text{v}_1-\text{v}_2}$
$\therefore$ Beat frequency $=\text{v}_1-\text{v}_2$
$\therefore$ no. of beats formed per sec. $=\text{v}_1-\text{v}_2$
- If R is minimum, then
$\cos\pi(\text{v}_1-\text{v}_2)\text{t}=\text{min}=0=\cos(2\text{n}+1)\frac{\pi}{2}$
$\therefore \pi(\text{v}_1-\text{v}_2)\text{t}=(2\text{n}+1)\frac{\pi}{2}$
$\text{t}=\frac{(2\text{n}+1)}{2(\text{v}_1-\text{v}_2)},$ where n = 0, 1, 2,...
$\therefore$ Amplitude becomes minimum at times grven by
$\text{t}=\frac{1}{2(\text{v}_1-\text{v}_2)},\frac{3}{2(\text{v}_1-\text{v}_2)},\frac{5}{2(\text{v}_1-\text{v}_2)},...$
$\therefore$ Time interval between two consecutive minima is $=\frac{1}{\text{v}_1-\text{v}_2}.$
$\therefore$ Beat period $=\frac{1}{\text{v}_1-\text{v}_2}.$
$\therefore$ Beat requency = v1 - v2.
$\therefore$ No. of beats formed per sec = v1 - v2.
Hence the number of beats formed per second is equal to the difference between the frequencies of two component waves.Need a full question paper?
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