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?

200 Hz

202 Hz

196 Hz

204 HN

Answer

196 Hz

|f₁−f₂| =4 Since mass of second tuning fork increases so f₂ decrease and beats increase so f₁>f₂ ⇒ f₂=f₁−4 = 196