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?

(1) 200 Hz

(2) 202 Hz

(3) 196 Hz

(4) 204 Hz

Answer 1

The correct answer is (3)

|f₁−f₂| =4

Since mass of second tuning fork increases so f₂ decrease and beats increase so

f₁>f₂
⇒ f₂=f₁−4 = 196