Two sound waves $y_{1} = 4 sim 604 π (t - \frac{x}{330})$ and $y_{2} = 3 \sin 596 π (t - \frac{x}{330})$ are superimposed and produces beats. Match I and II

I II
A) Beat frequency P) 49
B) Frequency of variation of amplitude Q) 4Hz
C) Frequency of resultant wave R) 2Hz
D) Ratio of maximum and minimum intensity of beats S) 300Hz
A B C D
1 Q R S p
2 Q S R P
3 R S P Q
4 P Q R S

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{aligned} ω_{1} = 2 {πn}_{1} = 604 π ∴ n_{1} = 302 \\ ω_{2} = 2 {πn}_{2} = 596 π ∴ n_{2} = 298 \end{aligned}$
A) Beat freq = $n_{1} - n_{2} = 4$
B) Freq. of amp. of variation $= \frac{n_{1} - n_{2}}{2} = 2$
C) Freq .of resultant wave $= \frac{n_{1} + n_{2}}{2} = 300$
D) $\frac{{I_{m ax}}_{}}{I_{min}} = {(\frac{A_{1} + A_{2}}{A_{1} - A_{2}})}^{2} = {(\frac{4 + 3}{4 - 3})}^{2} = 49$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

	I		II
A)	Beat frequency	P)	49
B)	Frequency of variation of amplitude	Q)	4Hz
C)	Frequency of resultant wave	R)	2Hz
D)	Ratio of maximum and minimum intensity of beats	S)	300Hz