Question

Two waves of frequency 250 Hz and 255 Hz are produced simultaneously. The time interval between successive maxima is

Answer 1

When two waves are produced simultaneously, having slightly different frequencies in the same direction superpose upon each other. If the difference between the two frequencies of waves is less than 10 than beats are produced. We can use the condition of beats frequency to derive the time interval between successive maxima.

Complete step by step answer:
To begin with let us first understand what does beat frequency mean. As mentioned in the hint if two waves have slightly difference in frequencies superpose each other as a result the intensity of the sound i.e. the amplitude in the medium rises and falls. One rise and one fall constitutes a beat. The number of beats heard per second is called the beat frequency.
Let us not derive the condition of beats but, let us obtain an expression between the maximum amplitude and time interval so that we can subtract the two and find the time interval between them.
Let us consider two waves of frequencies ${{v}_{1}}\text{ and }{{v}_{2}}$. The displacements of the two waves at a given point is given by,
Wave 1 ${{y}_{1}}=A\operatorname{Sin}{{\omega }_{1}}t=A\operatorname{Sin}2\pi {{v}_{1}}t$ and wave 2 ${{y}_{2}}=A\operatorname{Sin}{{\omega }_{2}}t=A\operatorname{Sin}2\pi {{v}_{2}}t$ , where A is the amplitude of the wave, ${{\omega }_{1}}\text{ and }{{\omega }_{2}}$ are the angular frequencies and since $\omega$is equal to $2\pi v$ where v is the frequency after substituting we get the displacement written above.
By the principle of superposition of the two waves the resultant displacement is given by,
$y={{y}_{1}}+{{y}_{2}}$
$A\operatorname{Sin}2\pi {{v}_{1}}t+A\operatorname{Sin}2\pi {{v}_{2}}t.......(1)$
Taking A common we can write equation 1 as
$A(\operatorname{Sin}2\pi {{v}_{1}}t+\operatorname{Sin}2\pi {{v}_{2}}t)$
Using the trigonometric formula i.e. $\operatorname{Sin}A+\operatorname{Sin}B=2\operatorname{Cos}\left( \dfrac{A-B}{2} \right)\operatorname{Sin}\left( \dfrac{A+B}{2} \right)$ the above expression becomes,
$A(\operatorname{Sin}2\pi {{v}_{1}}t+\operatorname{Sin}2\pi {{v}_{2}}t)$

$2A\operatorname{Cos}2\pi \left( \dfrac{{{v}_{1}}-{{v}_{2}}}{2} \right)t.\operatorname{Sin}2\pi \left( \dfrac{{{v}_{1}}+{{v}_{2}}}{2} \right)$
Substituting $\left( \dfrac{{{v}_{1}}-{{v}_{2}}}{2} \right)={{V}_{1}}\text{ and }\left( \dfrac{{{v}_{1}}+{{v}_{2}}}{2} \right)={{V}_{2}}$ for simplicity the above equation becomes,
$2A\operatorname{Cos}(2\pi {{V}_{1}}t).\operatorname{Sin}(2\pi {{V}_{2}}t)$
Again substituting R=$2A\operatorname{Cos}(2\pi {{V}_{1}}t)$ for simplicity the above equation becomes,
$y=R.\operatorname{Sin}(2\pi {{V}_{2}}t)....(2)$ .
If we look at equation 2 the term R represents the amplitude of the wave. Now let us obtain the maximum values for R.
$R=2A\operatorname{Cos}(2\pi {{V}_{1}}t)$. The value of R will be maximum if $2\pi {{V}_{1}}t=\pm 1$. let us substitute ${{V}_{1}}$in the equation i.e.
$2\pi {{V}_{1}}t=\pm 1$
$2\pi \left( \dfrac{{{v}_{1}}-{{v}_{2}}}{2} \right)t=\pm 1$
$\pi ({{v}_{1}}-{{v}_{2}})t=n\pi$ hence t for maximum amplitude is, $t=\dfrac{1}{({{v}_{1}}-{{v}_{2}})}$. For n =1, $t=\dfrac{1}{({{v}_{1}}-{{v}_{2}})}$, for n=2, $t=\dfrac{2}{({{v}_{1}}-{{v}_{2}})}$
Subtracting the above consequent maxima we get, the time period of two successive maximas.
$\dfrac{2}{({{v}_{1}}-{{v}_{2}})}-\dfrac{1}{({{v}_{1}}-{{v}_{2}})}$
$\dfrac{1}{({{v}_{1}}-{{v}_{2}})}\sec$
Substituting the frequency of the source we get,
${{v}_{1}}=255Hz\text{ and }{{v}_{2}}=250Hz$
$\dfrac{1}{({{v}_{1}}-{{v}_{2}})}\sec$
$\dfrac{1}{(255-250)}\sec$

Hence the answer to the above question is $\dfrac{1}{5}\sec =0.2\sec$.

Note:
If the difference in the frequency is more than 10, beats will be produced. But the number of beats will be more than 10 which is not possible for the human ear to feel this difference in intensity of sound. The Expression for beats frequency is ${{v}_{BEAT}}={{v}_{1}}-{{v}_{2}}$ since $\text{time period}=\dfrac{\text{1}}{\text{frequency}}$.