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?
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?
Show Hint
- $\theta_{1} l_{2}^{2}=\theta_{2} l_{1}^{2}$
- $\theta_{1} l_{1}=\theta_{2} l_{2}$
- $\theta_{1} l_{1}^{2}=\theta_{2} l_{2}^{2}$
- $\theta_{1} l_{2}=\theta_{2} l_{1}$
The Correct Option is D
Approach Solution - 1
2. Equating angular accelerations: \[ \frac{g}{l_1} \theta_1 = \frac{g}{l_2} \theta_2 \] \[ \theta_1 l_2 = \theta_2 l_1 \] Therefore, the correct answer is (4) $\theta_{1} l_{2}=\theta_{2} l_{1}$.
Approach Solution -2
We are given two simple pendulums having lengths \( l_1 \) and \( l_2 \), with angular displacements \( \theta_1 \) and \( \theta_2 \), respectively. It is given that both pendulums have the same angular acceleration. We need to find the correct relationship between their displacements and lengths.
Concept Used:
For a simple pendulum performing small oscillations, the restoring torque per unit moment of inertia gives the angular acceleration:
\[ \alpha = -\frac{g}{l} \sin \theta \]For small angles (in radians), \( \sin \theta \approx \theta \). Thus, the angular acceleration can be approximated as:
\[ \alpha = -\frac{g}{l} \theta \]Step-by-Step Solution:
Step 1: Write angular acceleration for the two pendulums.
\[ \alpha_1 = -\frac{g}{l_1} \theta_1, \quad \alpha_2 = -\frac{g}{l_2} \theta_2 \]Step 2: Given that the angular accelerations are equal in magnitude.
\[ \alpha_1 = \alpha_2 \]Step 3: Substitute the expressions for \( \alpha_1 \) and \( \alpha_2 \).
\[ -\frac{g}{l_1} \theta_1 = -\frac{g}{l_2} \theta_2 \]Step 4: Simplify the equation by canceling \( g \) and the negative signs.
\[ \frac{\theta_1}{l_1} = \frac{\theta_2}{l_2} \]Final Computation & Result:
The correct relation between angular displacements and lengths is:
\[ \boxed{\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}} \]Final Answer: \( \dfrac{\theta_1}{l_1} = \dfrac{\theta_2}{l_2} \)
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