MCQ
A $1 cm$ long string vibrates with fundamental frequency of $256\, Hz$. If the length is reduced to $\frac{1}{4}cm$ keeping the tension unaltered, the new fundamental frequency will be
- A$64$
- B$256$
- C$512$
- ✓$1024$
✓
Answer
Correct option: D.
$1024$
d
(d) $n \propto \frac{1}{l}$==> $\frac{{{n_2}}}{{{n_1}}} = \frac{{{l_1}}}{{{l_2}}}$
(d) $n \propto \frac{1}{l}$==> $\frac{{{n_2}}}{{{n_1}}} = \frac{{{l_1}}}{{{l_2}}}$
==> ${n_2} = \frac{{{l_1}}}{{{l_2}}}{n_1} = \frac{{1 \times 256}}{{1/4}} = 1024Hz$
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