An unknown tuning fork gives 6 beats per second with a 384 Hz standard fork. Wax is added to the unknown fork, lowering its frequency, and the beat frequency decreases to 4 beats per second. If its original frequency is N Hz, find N.
Physics · Wave Motion · beat and properties of sound
JEE Mains NEET JEE Advanced

Question

An unknown tuning fork gives 6 beats per second with a 384 Hz standard fork. Wax is added to the unknown fork, lowering its frequency, and the beat frequency decreases to 4 beats per second. If its original frequency is N Hz, find N.

Physics · Wave Motion · beat and properties of sound

G1 Physics · Accepted Answer

Initially, |f−384| = 6. Therefore, f = 378 Hz or 390 Hz. Wax lowers the frequency. Since the beat frequency decreases, the frequency must move towards 384 Hz. Therefore the unknown fork was initially above 384 Hz: f = 390 Hz. Hence, N = 390.