Question

15, Two identical tuning forks vibrate at 413 Hz. A small piece of clay is placed on one of tuning forks resulting in ten beats per second being heard. Calculate the period of the altered tuning fork a) 2.42 X 10-3s b) 2.48 X 10 s c) 2.54 X 10's d) 2.60 X 10s e) 2.36 X 103
The answer is B

Answer 1

Clay will make the fork vibrate more slowly. So, tuning fork settle on low frequency Beat frequency (F_b) = F_2 - F_1 10 = Vertical-bar 413 - F_1 Vertical-bar F_1 = 403 or 423 Hz But as per above mentioned we reject 423Hz F_1 = 403Hz (T = 1/f_1) Time period = 1/403 = 2.48 times 10^- 3 Hz Option B