Sound WavesMedium
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?
Options
A.200 Hz
B.202 Hz
C.196 Hz
D.204 Hz
Solution
|f1 - f2| = 4
Since mass of second tuning fork increases so f2 decrease and beats increase so
f1 > f2
⇒ f2 = f1 - 4 = 196
Since mass of second tuning fork increases so f2 decrease and beats increase so
f1 > f2
⇒ f2 = f1 - 4 = 196
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