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 $\frac{t_1^2}{t_2^2}$ is

• A. $2$
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $\sqrt{2}$

FBD from conical pendulum :


$T\cos \theta =mg=$ constant
$\Rightarrow T_1\cos {\theta }_1=T_2\cos {\theta }_2$
$\frac{T_1}{T_2}=\frac{\cos {\theta }_2}{\cos {\theta }_1}=\frac{\cos {60}^{\circ }}{\cos {45}^{\circ }}$
$=\frac{1/2}{1/\sqrt{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow {\left(\frac{T_1}{T_2}\right)}^2=\frac{1}{2}$

Along radial direction,
$T_1\sin {\theta }_1=m{\omega }_1^2{r}_1$ and
$T_2\sin {\theta }_2=m{\omega }_2^2{r}_2$
$\Rightarrow \frac{{\omega }_1^2}{{\omega }_2^2}=\frac{T_1\sin {\theta }_1}{T_2\sin {\theta }_2}\times \frac{r_2}{r_1}$
Also from question diagram,
$r_1=l\sin {\theta }_1\&r_2=l\sin {\theta }_2$ where $l$ is the length of the string.
$\Rightarrow \frac{r_1}{r_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}$

$\therefore \frac{{\omega }_1^2}{{\omega }_2^2}=\frac{T_1}{T_2}$ and since time period $t=\frac{2\pi }{\omega }$,
$\frac{t_1}{t_2}=\frac{{\omega }_2}{{\omega }_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{2}$