Question

The frequency of a tuning fork $A$ is $2\%$ greater than that of a standard fork ‘$K$'. The frequency of another tuning fork ‘$B$’ is $3\%$ less than ‘$K$’. When ‘$A$’ and ‘$B$’ are vibrated together $6$ beats per second are heard. The frequencies of ‘$A$’ and ‘$B$’ are:

• A. 122.4Hz,116.4Hz
• B. 132.4Hz,116.4Hz
• C. 142.4Hz,116.4Hz
• D. 152.4Hz,116.4Hz

Answer 1

Answer: (A)

122.4Hz,116.4Hz

Frequency of A $=\frac{102}{100}k$
Frequency of B $=\frac{97}{100}k$

$Beat=FrequencyofA-FrequencyofB$
$6=\frac{102}{100}k-\frac{97}{100}$
$6=\frac{5}{100}k$
$k=\frac{600}{5}=120H_z$

So, frequency of A $=\frac{102}{100}k=\frac{102}{100}\times 120$$=122.4H_z$
Frequency of B$=\frac{97}{100}k=\frac{97}{100}\times 120$$=116.4H_z$