Question

The frequency of a tuning fork P is $a%$ less than a standard fork A. The frequency of another fork Q is $b%$ greater than that of A. When P and Q are sounded together, $x$ beats are produced in one second. The frequency of the standard fork is (in Hz)

• A. $\dfrac{100x}{a+b}$
• B. $\dfrac{100x}{a-b}$
• C. $\dfrac{100x}{b-a}$
• D. $\dfrac{200x}{b-a}$

Hint:

Beat frequency is always equal to the absolute difference between the frequencies of the two sources.

Answer 1

The Correct Option is A

Step 1: Let the frequency of the standard fork A be $f$ Hz.
Step 2: Frequency of fork P (which is $a%$ less than A): \[ f_P = f\left(1-\frac{a}{100}\right) \]
Step 3: Frequency of fork Q (which is $b%$ greater than A): \[ f_Q = f\left(1+\frac{b}{100}\right) \]
Step 4: Number of beats per second equals the absolute difference of frequencies: \[ x = |f_Q - f_P| \]
Step 5: Substitute the expressions: \[ x = f\left(\frac{a+b}{100}\right) \]
Step 6: Solve for $f$: \[ f = \frac{100x}{a+b} \]