Question

A tuning fork $A$ produces $4$ beats with tuning fork $B$ of frequency $256Hz$. When $A$ is filed beats are found to occur at shorter intervals. What was its original frequency of tuning fork $A$ ?

• A. 260 Hz
• B. 252 Hz
• C. 256 Hz
• D. 258 Hz

Answer 1

Answer: (A)

As tuning fork $A$ produces $4$ beats with tuning fork $B$ of frequency $v_B(=256Hz)$, the frequency of $A\left(v_A\right)$ will be
$v_A=v_B\pm 4=256\pm 4,\text{ i.e., }{v}_A=252Hz$ or $260Hz$
Now on filing due to decrease in inertia frequency of $A$ will increase and occurrence of beats at shorter duration means increase in beat frequency; so if
$v_A=252Hz$, then
$256-v_A=4Hz$
and so with increase in $v_A$ beat frequency will decrease.
If $v_A=260Hz$, then
$v_A-256=4Hz$
and so with increase in $v_A$ beat frequency will increase and as on filing beat frequency increases the frequency of $A$ before filing was $260Hz$.