6. (AIEEE 2009)

Three sound waves of equal amplitudes have frequencies $(v - 1), v, (v + 1) .$ They superpose to give beats. The number of beats produced per second will be :

A. 3

B. 2

C. 1

D. 4

Correct Answer is Option (B)

Maximum number of beats $= (v + 1) - (v - 1) = 2$

7. (AIEEE 2005)

When two tuning forks (fork $1$ and fork $2$ ) are sounded simultaneously, $4$ beats per second are heated. Now, some tape is attached on the prong of the fork $2.$ When the tuning forks are sounded again, $6$ beats per second are heard. If the frequency of fork $1$ is $200$ $H z$ , then what was the original frequency of fork $2$ ?

A. 202 Hz

B. 200 Hz

C. 204 Hz

D. 196 Hz

Correct Answer is Option (D)

No. of beats heard when fork $2$ is sounded with fork $1$ $= Δ n = 4$

Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from $4$ to $6$ in this case) then the frequency of the unknown fork $2$ is given by,

$n = n_{0} - Δ n = 200 - 4 = 196 H z$

8. (AIEEE 2003)

A tuning fork of known frequency $256$ $H z$ makes $5$ beats per second with the vibrating string of a piano. The beat frequency decreases to $2$ beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

A. $256 + 2 H z$

B. $256 - 2 H z$

C. $256 - 2 H z$

D. $256 + 5 H z$

Correct Answer is Option (C)

A tuning fork of frequency $256$ $H z$ makes $5$ beats/ second with the vibrating string of a piano. Therefore the frequency of the vibrating string of piano is $(256 \pm 5)$ $H z$ ie either $261$ $H z$ or $251$ $H z .$ When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is $251$ $H z$

9. (AIEEE 2002)

A tuning fork arrangement (pair) produces $4$ beats/sec with one fork of frequency $288$ $c p s .$ A little wax is placed on the unknown fork and it then produces $2$ beats/sec. The frequency of the unknown fork is :

A. 286 cps

B. 292 cps

C. 294 cps

D. 288 cps

Correct Answer is Option (B)

A tuning fork produces $4$ beats/sec with another tuning fork of frequency $288$ cps. From this information we can conclude that the frequency of unknown fork is $288 + 4$ $c p s$ or $288 - 4$ $c p s$ i.e. $292$ $c p s$ or $284$ $c p s .$

Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4. So the difference between frequency decreases.

This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.

So the frequency of the unknown tuninh fork is = 292 Hz

Waves