Question

When the string of a conical pendulum makes an angle of $45^{\circ }$ with the vertical, its time period is $T_1$. When the string makes an angle of $60^{\circ }$ with the vertical, its time period is $T_2$. Then $T_1^2/T_2^2$ is

• A. $2$
• B. $\sqrt{2}$
• C. $0.5$
• D. none of these

Answer 1

Answer: (B)

$<br/>\begin{array}{l}<br/>T\cos \theta =mg=\text{ constant }\\ <br/>\Rightarrow T_1\cos {\theta }_1=T_2\cos {\theta }_2\\ <br/>\frac{T_1}{T_2}=\frac{\cos {\theta }_2}{\cos {\theta }_1}\frac{\cos 60}{\cos 45}\\ <br/>=\frac{1/2}{1/\sqrt{2}}=\frac{1}{\sqrt{2}}\\ <br/>\Rightarrow {\left(\frac{T_1}{T_2}\right)}^2=\frac{1}{2}<br/>\end{array}<br/>$
Along radial direction,
$<br/>\begin{array}{l}<br/>T_1\sin {\theta }_1=m{\omega }_1^2{r}_1\text{ and }\\ <br/>T_2\sin {\theta }_2=m{\omega }_2^2{r}_2\\ <br/>\Rightarrow \frac{{\omega }_1^2}{{\omega }_2^2}=\frac{T_1\sin {\theta }_1}{T_2\sin {\theta }_2}\times \frac{r_2}{r_1}<br/>\end{array}<br/>$
Also from question diagram,
$r_1=\left|\sin {\theta }_1\&r_2=\right|\sin {\theta }_2$ where $\mid$ is the length of the string.
$<br/>\Rightarrow \frac{r_1}{r_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}<br/>$
$\therefore \frac{{\omega }_1^2}{{\omega }_2^2}=\frac{T_1}{T_2}$ and since time period $t=\frac{2\pi }{\omega }$,
$<br/>\frac{t_1}{t_2}=\frac{{\omega }_2}{{\omega }_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{2}<br/>$