∣f1−f2∣=4 Since mass of second tuning fork increases so f2 decrease and beats increase so f1>f2⇒f2=f1−4=196
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 ?
Held on 30 Apr 2005 · Verified 6 Jul 2026.
200 Hz
202 Hz
196 Hz
204 Hz
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