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  • A simple Pendulum has time Period  T1=12 sec. The Point of suspension is now moved upward according to the relation <img alt="Y=K t^{^{2}}" src="https://entrancecorner.oncodeco

A simple Pendulum has time Period  T1=12 sec. The Point of suspension is now moved upward according to the relation Y=K t^{^{2}}  (K=2 m/s^{2}), where Y  is the vertical displacement. The time Period  T2=14  sec. The ratio of  \frac{l_1}{l_2}=?  (Take  g=10 \mathrm{~m} / \mathrm{s}^2)

 

Option: 1

\frac{l_1}{l_2}=\frac{150}{413}


Option: 2

\frac{l_1}{l_2}=\frac{179}{354}


Option: 3

\frac{l_1}{l_2}=\frac{189}{353}


Option: 4

\frac{l_1}{l_2}=\frac{180}{343}


Answers (1)

best_answer

Y=K t^2 \\

a=\frac{d^2 Y}{d t^2}=2 K \quad[\because K=2] \\

a=2 \times 2 \\

a=4 \mathrm{~m} / \mathrm{s}^2 \\

\text { geff }=a+g=4+10=14 \mathrm{~m} / \mathrm{s}^2

T_1=2 \pi \sqrt{\frac{l_1}{g}} \\

12=2 \pi \sqrt{\frac{l_1}{g}} \\
or=12=2 \pi \sqrt{\frac{l_1}{10}}\rightarrow eq(1)

T_2=2 \pi \sqrt{\frac{l_2}{14}} \\

14=2 \pi \sqrt{\frac{l_2}{14}}\rightarrow eq(2)

dividing (1) by (2)

\frac{12}{14}=\frac{2 \pi \frac{\sqrt{l_1}}{\sqrt{g}}}{2 \pi \frac{\sqrt{l _2}}{\sqrt{g}}} \\

\frac{12}{14}=\frac{\sqrt{l_1}}{\sqrt{g}} \times \frac{\sqrt{ {g}'}}{\sqrt{l^2}} \\

\frac{12}{14}=\frac{\sqrt{l _1}}{\sqrt{l _2}} \times \frac{\sqrt{14}}{\sqrt{10}}

\frac{12 \sqrt{10}}{14 \sqrt{14}}=\frac{\sqrt{l_1}}{\sqrt{l_2}} \\

\frac{6}{7} \sqrt{\frac{10}{14}}=\frac{\sqrt{l_1}}{\sqrt{l_2}} \\

\frac{36}{49} \times \frac{10}{14}=\frac{l_1}{l_2} \\

\frac{36 \times 5}{7 \times 49}=\frac{l_1}{l_2} \\

\frac{l_1}{l_2}=\frac{180}{343}

Posted by

Deependra Verma

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