If $n_{1}$ and $n_{2}$ are the frequencies of sound waves producing beats, then the time interval between one maxima and next minima is ?

\frac{1}{n_{1} + n_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{n_{1} - n_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2 (n_{1} + n_{2})}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2 (n_{1} - n_{2})}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{1}{2 (n_{1} - n_{2})}$
Given, two difference frequencies $n_{1} a n d n_{2}$ .We have to find time difference between one minima and maxima.

We know that
$f_{b e a t} = n_{1} - n_{2}$

So, $T_{b e a t} = \frac{1}{f_{b e a t}} = \frac{1}{(n_{1} - n_{2})}$

Between one maxima and minima, the time difference is $\frac{T_{b e a t}}{2}$

So, $T_{r e q} = \frac{1}{2 (n 1 - n 2)}$

So, D is the correct answer.

Suggest Corrections

Similar questions

Q. When two waves of almost equal frequency

n_{1}

and

n_{2}

are produced simultaneously, then the times interval between successive maxima is

Q. If

n_{1}, n_{2}

and

n_{3}

are the fundamental frequencies of three segments of a string of length

l

, then the original fundamental frequency

n

of the string is given by

If $n_{1}, n_{2}$ and $n_{3}$ are the fundamental frequencies of three segments into which a string is divided, then the fundamental frequency n of the original string is given by

Q. Two tuning forks of frequency

n_{1}

and

n_{2}

produces n beats per second. If

n_{2}

and

n

are known,

n_{1}

may be given by :

Q. Let

N_{1} = 2^{55} + 1

and

N_{2} = 165.

Then

Join BYJU'S Learning Program

If n1 and n2 are the frequencies of sound waves producing beats, then the time interval between one maxima and next minima is ?

If $n_{1}$ and $n_{2}$ are the frequencies of sound waves producing beats, then the time interval between one maxima and next minima is ?