The frequency of a tuning fork A is 2 % more than the frequency of a standard tuning fork. The frequency of another fork B is 3 % less than the frequency of a standard tuning fork. If 6 beats s- 1 are heard when the two tuning forks A and B are excited, the frequency of A is

Question

The frequency of a tuning fork A is 2 % more than the frequency of a standard tuning fork. The frequency of another fork B is 3 % less than the frequency of a standard tuning fork. If 6 beats s- 1 are heard when the two tuning forks A and B are excited, the frequency of A is (A) 120Hz (B) 122.4Hz (C) 116.4Hz (D) 130Hz

Tardigrade Admin · Accepted Answer

Let the frequency of standard fork =x ∴ nA=(102/100)x , nB=(97/100)x Number of beats per second =nA-nB=6 (102/100)x-(97/100)x=6 x=(6 × 100/5)=120 Hz ∴ nA=(102/100)x=122.4Hz

Q. The frequency of a tuning fork $A$ is $2%$ more than the frequency of a standard tuning fork. The frequency of another fork $B$ is $3%$ less than the frequency of a standard tuning fork. If $6 b e a t s s^{- 1}$ are heard when the two tuning forks $A$ and $B$ are excited, the frequency of $A$ is

$120 Hz$

$122.4 Hz$

$116.4 Hz$

$130 Hz$

Let the frequency of standard fork $= x$
$∴ n_{A} = \frac{102}{100} x$ , $n_{B} = \frac{97}{100} x$
Number of beats per second $= n_{A} - n_{B} = 6$
$\frac{102}{100} x - \frac{97}{100} x = 6$
$x = \frac{6 \times 100}{5} = 120 Hz$
$∴ n_{A} = \frac{102}{100} x = 122.4 Hz$

Q. The frequency of a tuning fork A is 2% more than the frequency of a standard tuning fork. The frequency of another fork B is 3% less than the frequency of a standard tuning fork. If 6beatss−1 are heard when the two tuning forks A and B are excited, the frequency of A is

120Hz

122.4Hz

116.4Hz

130Hz