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?
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200 Hz
-
202 Hz
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196 Hz
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204 HN
Solution
C.
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
