Question:
A simple pendulum consists of a small bob of mass 40 g at the end of a cord of negligible mass. The angle $\theta$ between the cord and the vertical is given by $\theta = \theta_0 \cos(4t + \phi)$ rad. If the maximum kinetic energy is $1.25 \times 10^{-3}$ J, then $\theta_0$ is:
A simple pendulum consists of a small bob of mass 40 g at the end of a cord of negligible mass. The angle $\theta$ between the cord and the vertical is given by $\theta = \theta_0 \cos(4t + \phi)$ rad. If the maximum kinetic energy is $1.25 \times 10^{-3}$ J, then $\theta_0$ is:
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In pendulum SHM, always convert angular motion to energy form using $KE_{\max} = \frac{1}{2} I \omega^2 \theta_0^2$.
Updated On: Jul 18, 2026
- 0.08 rad
- 1.75 rad
- 4.25 rad
- 0.1 rad
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The Correct Option is D
Solution and Explanation
Step 1: Understanding SHM of a pendulum.
The given equation $\theta = \theta_0 \cos(4t+\phi)$ represents angular SHM of a simple pendulum. Here angular frequency is: \[ \omega = 4 \, \text{rad s}^{-1} \] For a simple pendulum, $\omega^2 = \frac{g}{l}$, which connects geometry with dynamics.
Step 2: Finding length of pendulum using angular frequency.
Using: \[ \omega^2 = \frac{g}{l} \Rightarrow l = \frac{g}{\omega^2} \] Substituting $g=10$ and $\omega=4$: \[ l = \frac{10}{16} = 0.625 \, \text{m} \] This step is crucial because energy expression depends on moment of inertia.
Step 3: Expression for maximum kinetic energy.
For small oscillations, maximum kinetic energy equals total energy: \[ KE_{\max} = \frac{1}{2} m l^2 \omega^2 \theta_0^2 \] This comes from rotational SHM where $I = ml^2$.
Step 4: Substituting numerical values.
Given: \[ m = 0.04 \, \text{kg}, \quad l = 0.625, \quad \omega = 4 \] So: \[ KE_{\max} = \frac{1}{2} \times 0.04 \times (0.625)^2 \times 16 \times \theta_0^2 \] Now simplifying step-by-step: \[ \frac{1}{2} \times 0.04 = 0.02 \] \[ (0.625)^2 = 0.390625 \] \[ 0.02 \times 0.390625 = 0.0078125 \] \[ 0.0078125 \times 16 = 0.125 \] So: \[ KE_{\max} = 0.125 \, \theta_0^2 \]
Step 5: Solving for $\theta_0$.
Given: \[ 1.25 \times 10^{-3} = 0.125 \theta_0^2 \] \[ \theta_0^2 = \frac{1.25 \times 10^{-3}}{0.125} = 10^{-2} \] \[ \theta_0 = 0.1 \, \text{rad} \]
Step 6: Final conclusion.
\[ \boxed{0.1 \, \text{rad}} \]
The given equation $\theta = \theta_0 \cos(4t+\phi)$ represents angular SHM of a simple pendulum. Here angular frequency is: \[ \omega = 4 \, \text{rad s}^{-1} \] For a simple pendulum, $\omega^2 = \frac{g}{l}$, which connects geometry with dynamics.
Step 2: Finding length of pendulum using angular frequency.
Using: \[ \omega^2 = \frac{g}{l} \Rightarrow l = \frac{g}{\omega^2} \] Substituting $g=10$ and $\omega=4$: \[ l = \frac{10}{16} = 0.625 \, \text{m} \] This step is crucial because energy expression depends on moment of inertia.
Step 3: Expression for maximum kinetic energy.
For small oscillations, maximum kinetic energy equals total energy: \[ KE_{\max} = \frac{1}{2} m l^2 \omega^2 \theta_0^2 \] This comes from rotational SHM where $I = ml^2$.
Step 4: Substituting numerical values.
Given: \[ m = 0.04 \, \text{kg}, \quad l = 0.625, \quad \omega = 4 \] So: \[ KE_{\max} = \frac{1}{2} \times 0.04 \times (0.625)^2 \times 16 \times \theta_0^2 \] Now simplifying step-by-step: \[ \frac{1}{2} \times 0.04 = 0.02 \] \[ (0.625)^2 = 0.390625 \] \[ 0.02 \times 0.390625 = 0.0078125 \] \[ 0.0078125 \times 16 = 0.125 \] So: \[ KE_{\max} = 0.125 \, \theta_0^2 \]
Step 5: Solving for $\theta_0$.
Given: \[ 1.25 \times 10^{-3} = 0.125 \theta_0^2 \] \[ \theta_0^2 = \frac{1.25 \times 10^{-3}}{0.125} = 10^{-2} \] \[ \theta_0 = 0.1 \, \text{rad} \]
Step 6: Final conclusion.
\[ \boxed{0.1 \, \text{rad}} \]
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