A string fixed at both ends is in resonance in its 2^ nd harmonic… - Vidyadip

MCQ

A string fixed at both ends is in resonance in its $2^{nd}$ harmonic with a tuning fork of frequency $f_1$. Now its one end becomes free. If the frequency of the tuning fork is increased slowly from $f_1$ then again a resonance is obtained when the frequency is $f_2$. If in this case the string vibrates in $n^{th}$ harmonic then

A
$n = 3, f_2 =\frac{3}{4}f_1$
B
$n = 3, f_2 =\frac{5}{4}f_1$
✓
$n = 5, f_2 =\frac{5}{4}f_1$
D
$n = 5, f_2 =\frac{3}{4}f_1$

✓

Answer

Correct option: C.

$n = 5, f_2 =\frac{5}{4}f_1$

c
$\mathrm{F}_{1}=\frac{2}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

$\mathrm{F}_{2}=\frac{5}{4} \mathrm{F}_{1}$

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