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 b

Question

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 seconds are heard. If the frequency of fork 1 is 200 Hz , then what was the original frequency of fork 2 ?

TestHub · Accepted Answer

∣ f 1 ​ − f 2 ​ ∣ = 4 Since mass of second tuning fork increases so f 2 ​ decrease and beats increase so ​ f 1 ​ > f 2 ​ ⇒ f 2 ​ = f 1 ​ − 4 = 196 ​

Physics - Sound Wave Question with Solution | TestHub

Options:

Doubts & Discussion