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14.Waves and Sound
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The frequency of transverse vibrations in a stretched string is $200 Hz$. If the tension is increased four times and the length is reduced to one-fourth the original value, the frequency of vibration will be .... $Hz$
A
$25$
B
$200$
C
$400 $
D
$1600$
Solution
(d) $n = \frac{1}{{2l}}\sqrt {\frac{T}{m}} $
==>$\frac{{{n_1}}}{{{n_2}}} = \frac{{{l_2}}}{{{l_1}}}\sqrt {\frac{{{T_1}}}{{{T_2}}}} $$ = \frac{1}{4}\sqrt {\frac{1}{4}} = \frac{1}{8}$
==> ${n_2} = 8{n_1} = 8 \times 200 = 1600Hz$
Standard 11
Physics