Test: Beats - JEE MCQ

To solve this problem, let's assume the original frequency of string B is x Hz.

When two slightly out-of-tune guitar strings produce beats, the beat frequency is equal to the difference in frequency between the two strings. In this case, the beat frequency is 6 Hz.

So, we can set up the following equation:

|430 Hz - x Hz| = 6 Hz

Now, we have two cases to consider:

1. If 430 Hz - x Hz = 6 Hz: In this case, the original frequency of string B would be 430 Hz - 6 Hz = 424 Hz.

Frequency of fork one is 346 Hz.
When the other fork is waxed, the beat is increased.
So, the frequency of other fork is less than the frequency of fork one.
So, 
beat = frequency of fork one − frequency of second fork
4=346− Frequency of second fork
Frequency of second fork =350 Hz

A tuning fork is made of an alloy of steel, nickel and chromium, called elinvar, i.e. the material for which the elasticity does not change.

Radar uses electromagnetic energy pulses. The radio-frequency (rf) energy is transmitted to and reflected from the reflecting object. A small portion of the reflected energy returns to the radar set. This returned energy is called an ECHO. Radar sets use the echo to determine the direction and distance of the reflecting object.
It does not work on phenomena of beats.

For the formation of distinct beats, frequencies of two sources of sound should be nearly equal, i.e., difference in frequencies of two sources must be small, say less than 10. The impression of sound heard by ow ears persists on our mind for 1 /10th of a second. If another sound is heard before (1 /10) second passes, the impressions of the two sounds mix up and our mind cannot distinguish between the two. In order to hear distinct beats, time interval between two successive beats must be greater than 1 /10 second. Therefore, frequency of beats must be less than 1 0,i.e., number of beats/sec, which is equal to difference in frequencies of two sources must be less than 10. Hence the two sources should be of nearly equal frequencies. 

According to the Principle of Superposition, if the particle of a medium is given an order (i.e. of oscillating sinusoidally) and then another order (similar or different), the particle simply adds the two orders vectorially (i.e. by vector method), and then oscillates by the amplitude of the resultant vector.
So if the particle of a medium is given two similar orders (i.e., both the sine waves having the same amplitude, say A) then the particle will oscillate with an amplitude, 2A, i.e., the amplitude will be doubled.

As we know time interval will have dimensions of time so check in option for dimension of time.
Dimensions of v is "per second" or [T^-1]

Number extra waves received from each side:

n = +- u/λ

where

u= speed of the observer

λ = Wavelength of sound

Number of beats:The number of beats is given by the difference between number of extra waves received on either sides.

Given that two sources produce pure sound notes as follows: y1= 0.008 sin (604 n t) y2= 0.007 sin (610 n t) To find the number of beats heard by a person at the location, we need to first understand the concept of beats.

Beats: When two sound waves of slightly different frequencies are superimposed, we hear a periodic variation in the loudness of sound. This variation is called beats.

The number of beats heard per second is given by the difference between the frequencies of the two sound waves.

Mathematically, the beat frequency is given by fbeat = |f1 - f2| where f1 and f2 are the frequencies of the two sound waves. In this problem, the frequencies of the two sound waves are 604 n and 610 n, respectively.

Therefore, the beat frequency is fbeat = |604 n - 610 n| = 6 n Since the beat frequency is 6 n, the number of beats heard per second is 6. Since each beat corresponds to two sound waves (one from each source), the total number of sound waves heard per second is twice the number of beats, which is 12.

However, the question asks for the number of distinct sounds heard, which is equal to the number of beats. Therefore, the answer is 3 (Option A).
 

The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 258 Hz and 256 Hz are played simultaneously, a beat frequency of 2 Hz will be detected.