Question 53: Oscillations

Question

Two simple pendulums having lengths $l_1$ and $l_2$ with negligible string mass undergo angular displacements $	heta_1$ and $	heta_2$, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?

Accepted Answer

Angular Frequency
: The angular frequency ($\omega$) of a simple pendulum is given by:
$    \omega = \sqrt{\frac{g}{\ell}}    $
where $g$ is the acceleration due to gravity and $\ell$ is the pendulum length.
Angular Acceleration
: The angular acceleration ($\alpha$) can be expressed in terms of angular displacement ($	heta$) and angular frequency:
$    \alpha = -\omega^2 	heta    $
Equating Angular Accelerations
: Since the angular accelerations of the two pendulums are equal, we equate them:
$    \frac{g}{\ell_1} 	heta_1 = \frac{g}{\ell_2} 	heta_2    $
Simplifying the Expression
: By canceling out $g$ on both sides, we derive:
$    	heta_1 \ell_2 = 	heta_2 \ell_1    $
Thus, the correct expression that relates the displacements and lengths of the pendulums is $	heta_1 \ell_2 = 	heta_2 \ell_1$.

Answer

$	heta_1 l_2=	heta_2 l_1$

Answer

$	heta_1 l_1=	heta_2 l_2$

Answer

$	heta_1 l_2^2=	heta_2 l_1^2$

Answer

$	heta_1 l_1^2=	heta_2 l_2^2$

Oscillations

Two simple pendulums having lengths l_1 and l_2 with negligible string mass undergo angular displacements \theta_1 and \theta_2, from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?

Choose the correct answer: