A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is _________ Hz.
Answer (integer)
152
Solution
<p>Given ${v_{20}} = 2{v_1}$</p>
<p>Also ${v_{20}} = 4 \times 19 + {v_1}$</p>
<p>So ${v_{20}} = 152\,Hz$</p>
About this question
Subject: Physics · Chapter: Waves · Topic: Wave Motion and Types
This question is part of PrepWiser's free JEE Main question bank. 88 more solved questions on Waves are available — start with the harder ones if your accuracy is >70%.