Waves

JEE Main level, single correct, medium

124 of 142

Question 124 of 142

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 Hz200 \mathrm{~Hz}, then what was the original frequency of fork 2 ?

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Answer: (C) 196 Hz196 \mathrm{~Hz}

∣f1−f2∣=4\left|f_1-f_2\right|=4
Since mass of second tuning fork increases so f2f_2 decrease and beats increase so
f1>f2⇒f2=f1−4=196\begin{aligned} & f_1>f_2 \\ & \Rightarrow f_2=f_1-4=196 \end{aligned}

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