wave_mechanics mock_test_waves_and_acoustics Question 14
Question: A tuning fork produces 4 beats per second with another fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats per second. The unknown frequency is
Options:
A) 286 cps
B) 284 cps
C) 292 cps
D) 290 cps
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The tuning fork of frequency 288 Hz is producing 4 beats per second with the unknown tuning fork, i.e., the frequency difference between them is 4.
Therefore, the frequency of the unknown tuning fork is 288 ± 4 = 292 or 284.
On placing a little wax on the unknown tuning fork, its frequency decreases but now the number of beats produced per second is 2, i.e., the frequency difference now decreases.
It is possible only when before placing the wax, the frequency of the unknown fork is greater than the frequency of the given tuning fork.
Hence, the frequency of the unknown tuning fork is 292 Hz.
BETA
