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When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?
  • a)
    202 Hz
  • b)
    200 Hz
  • c)
    204 Hz
  • d)
    196 Hz
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
When two tuning forks (fork 1 and fork 2) are sounded simultaneously, ...
No. of beats heard when fork 2 is sounded with fork 1 = Δn = 4
Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by,
n = n0 - Δn = 200 – 4 = 196 Hz
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When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?a)202 Hzb)200 Hzc)204 Hzd)196 HzCorrect answer is option 'D'. Can you explain this answer?
Question Description
When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?a)202 Hzb)200 Hzc)204 Hzd)196 HzCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?a)202 Hzb)200 Hzc)204 Hzd)196 HzCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?a)202 Hzb)200 Hzc)204 Hzd)196 HzCorrect answer is option 'D'. Can you explain this answer?.
Solutions for When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?a)202 Hzb)200 Hzc)204 Hzd)196 HzCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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