Beats are produced by two waves given by y_1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The number of beats heard per second is given by the difference in frequencies of the two waves,$|n_1 - n_2|$.Given the equations $y_1 = a \sin(20

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The number of beats heard per second is given by the difference in frequencies of the two waves,$|n_1 - n_2|$.Given the equations $y_1 = a \sin(2000 \pi t)$ and $y_2 = a \sin(2008 \pi t)$,we compare these with the standard form $y = a \sin(2 \pi n t)$.For the first wave: $2 \pi n_1 = 2000 \pi$,which gives $n_1 = 1000 	ext{ Hz}$.For the second wave: $2 \pi n_2 = 2008 \pi$,which gives $n_2 = 1004 	ext{ Hz}$.The number of beats per second is $|n_2 - n_1| = |1004 - 1000| = 4 	ext{ beats per second}$.

Beats are produced by two waves given by $y_1 = a \sin(2000 \pi t)$ and $y_2 = a \sin(2008 \pi t)$. The number of beats heard per second is

Similar Questions

What is a beat? Obtain the equation for the number of beats produced in unit time.

Two waves $y = 0.25 \sin(316t)$ and $y = 0.25 \sin(310t)$ are travelling in the same direction. The number of beats produced per second will be:

When temperature increases,the frequency of a tuning fork

$A$ source of frequency $\nu$ gives $6 \text{ beats/second}$ when sounded with a source of frequency $200 \text{ Hz}$. The second harmonic of frequency $2\nu$ of the source gives $8 \text{ beats/second}$ when sounded with a source of frequency $420 \text{ Hz}$. The value of $\nu$ is (in $text{ Hz}$)

Two waves $Y_1 = 0.25 \sin(316t)$ and $Y_2 = 0.25 \sin(310t)$ are propagating along the same direction. The number of beats produced per second is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module