Question 279: Oscillations

Question

Let ' $l_1$ ' be the length of simple pendulum. Its length changes to ' $l_2$ ' to increase the periodic time by $20 \%$. The ratio $\frac{l_2}{l_1}=$

Accepted Answer

$$\begin{aligned}
&\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}\
&	ext { As the periodic time increases by } 20 \% 	ext {, }\
&\begin{aligned}
& \mathrm{T}_2=\mathrm{T}_1+\frac{20}{100}=\frac{120}{100} \mathrm{~T}_1 \
& \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{l_2}{l_1}} \
& \left(\frac{\mathrm{~T}_2}{\mathrm{~T}_1}ight)^2=\frac{l_2}{l_1} \
& \left(\frac{120 \mathrm{~T}_1}{100 	imes \mathrm{T}_1}ight)^2=\frac{l_2}{l_1} \
& \frac{36}{25}=\frac{l_2}{l_1} \
& \frac{l_2}{l_1}=1.44
\end{aligned}
\end{aligned}$$

Answer

1.22

Answer

1.33

Answer

1.44

Answer

1.55

Oscillations

Let ' l_1 ' be the length of simple pendulum. Its length changes to ' l_2 ' to increase the periodic time by 20 \%. The ratio \frac{l_2}{l_1}=

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