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If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$, then the ratio of the final and initial fundamental frequencies is :
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The correct answer is:
$2: 1$
Fundamental frequency, $n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
Initial frequency, $n_1=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
New length, $l^{\prime}=l-\frac{l \times 40}{100}=0.6 l$
New tension, $T^{\prime}=T+\frac{T \times 44}{100}$
$=1.44 \mathrm{~T}$
New frequency $n_2=\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}$
$\therefore \quad \frac{n_2}{n_1}=\frac{\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}$
$\frac{n_2}{n_1}=\frac{l}{l^{\prime}} \sqrt{\left(\frac{T^{\prime}}{T}\right)}$
$=\frac{l}{0.6 l} \sqrt{\frac{1.44 T}{T}}$
$=\frac{1}{0.6} \times 1.2$
$=2$
$\therefore \quad n_2: n_1=2: 1$
Initial frequency, $n_1=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
New length, $l^{\prime}=l-\frac{l \times 40}{100}=0.6 l$
New tension, $T^{\prime}=T+\frac{T \times 44}{100}$
$=1.44 \mathrm{~T}$
New frequency $n_2=\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}$
$\therefore \quad \frac{n_2}{n_1}=\frac{\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}$
$\frac{n_2}{n_1}=\frac{l}{l^{\prime}} \sqrt{\left(\frac{T^{\prime}}{T}\right)}$
$=\frac{l}{0.6 l} \sqrt{\frac{1.44 T}{T}}$
$=\frac{1}{0.6} \times 1.2$
$=2$
$\therefore \quad n_2: n_1=2: 1$
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