MCQ
Two wires are in unison. If the tension in one of the wires is increased by $2\%, 5$ beats are produced per second. The initial frequency of each wire is .... $Hz$
- A$200$
- B$400 $
- ✓$500$
- D$1000$
✓
Answer
Correct option: C.
$500$
c
(c) $n \propto \sqrt T $ ==> $\frac{{\Delta n}}{n} = \frac{{\Delta T}}{{2T}}$
If tension increases by $ 2\%$, then frequency must increases by $1\%$.
If initial frequency ${n_1} = n$ then final frequency $n_2 -n_1 = 5$
==> $\frac{{101}}{{100}}n - n = 5$ ==> $n = 500Hz.$
Short trick : If you can remember then apply following formula to solve such type of problems.
Initial frequency of each wire $(n)$
$ = \frac{{{\rm{(Number}}\;{\rm{of}}\;{\rm{beats}}\;{\rm{heard}}\;{\rm{per}}\;{\rm{sec)}} \times {\rm{200}}}}{{{\rm{(per}}\;{\rm{centage}}\;{\rm{change}}\;{\rm{in}}\;{\rm{tension}}\;{\rm{of}}\;{\rm{the}}\;{\rm{wire)}}}}$
Here $n = \frac{{5 \times 200}}{2} = 500Hz$
(c) $n \propto \sqrt T $ ==> $\frac{{\Delta n}}{n} = \frac{{\Delta T}}{{2T}}$
If tension increases by $ 2\%$, then frequency must increases by $1\%$.
If initial frequency ${n_1} = n$ then final frequency $n_2 -n_1 = 5$
==> $\frac{{101}}{{100}}n - n = 5$ ==> $n = 500Hz.$
Short trick : If you can remember then apply following formula to solve such type of problems.
Initial frequency of each wire $(n)$
$ = \frac{{{\rm{(Number}}\;{\rm{of}}\;{\rm{beats}}\;{\rm{heard}}\;{\rm{per}}\;{\rm{sec)}} \times {\rm{200}}}}{{{\rm{(per}}\;{\rm{centage}}\;{\rm{change}}\;{\rm{in}}\;{\rm{tension}}\;{\rm{of}}\;{\rm{the}}\;{\rm{wire)}}}}$
Here $n = \frac{{5 \times 200}}{2} = 500Hz$
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