A tuning fork gives $5$ beats with another tuning fork of frequency $100\,Hz.$ When the first tuning fork is loaded with wax, then the number of beats remains unchanged, then what will be the frequency of the first tuning fork ..... $Hz$
Medium
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(c) Suppose $n_A$ = known frequency $= 100\, Hz,$ $n_B = ?$
$x = 5\, bps,$ which remains unchanged after loading
Unknown tuning fork is loaded so $n_B\downarrow$
Hence $n_A -n_B \downarrow = x $... $(i)$ $\rightarrow$ correct
$n_B \downarrow -n_A = x $... $(ii)$ $\rightarrow$ wrong
From equation $ (i),$ it is clear that as $n_B$ decreases, beat frequency. (i.e.$ n_A -(n_B)_{new}$) can never be $x$ again.
From equation $(ii),$ as$ n_B\downarrow$, beat frequency (i.e. $(n_B)_{new} -n_A$) decreases as long as $(n_B)_{new}$ remains greater than $n_A$, If $(n_B)_{new}$ become lesser than $n_A$ the beat frequency will increase again and will be $x.$
$x = 5\, bps,$ which remains unchanged after loading
Unknown tuning fork is loaded so $n_B\downarrow$
Hence $n_A -n_B \downarrow = x $... $(i)$ $\rightarrow$ correct
$n_B \downarrow -n_A = x $... $(ii)$ $\rightarrow$ wrong
From equation $ (i),$ it is clear that as $n_B$ decreases, beat frequency. (i.e.$ n_A -(n_B)_{new}$) can never be $x$ again.
From equation $(ii),$ as$ n_B\downarrow$, beat frequency (i.e. $(n_B)_{new} -n_A$) decreases as long as $(n_B)_{new}$ remains greater than $n_A$, If $(n_B)_{new}$ become lesser than $n_A$ the beat frequency will increase again and will be $x.$
Hence this is correct.
So, $n_B = n_A + x = 100 + 5 = 105 Hz.$
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