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14.Waves and Sound
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If the length of stretched string is shortened by $40\%$ and the tension is increased by $44\%$, then the ratio of the final and initial fundamental frequencies is
A
$3 : 4$
B
$4 : 3$
C
$1 : 3$
D
$2 : 1$
Solution
$\mathrm{n}=\frac{1}{2} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
$\therefore \frac{{{n_2}}}{{{n_1}}} = \frac{{{1_1}}}{1}\sqrt {\frac{{{T_2}}}{{{T_1}}}} $
$=\frac{1_{1}}{\left[1_{1}-\frac{40}{100} 1_{1}\right]} \sqrt{\left(\frac{\mathrm{T}_{1}+\frac{44}{100} \mathrm{T}_{1}}{\mathrm{T}_{1}}\right)}$
$=\frac{100}{60} \times \frac{12}{10}=2: 1$
Standard 11
Physics
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