For a particle executing simple harmonic motion (SHM), let's analyze the situation at the mean position (equilibrium point).
The correct answer is (B) Velocity is maximum and acceleration is zero.
In simple harmonic motion (SHM), at the mean position (x = 0), the velocity is at its maximum and the acceleration is zero. The equations for SHM are: \[ V = \omega A \sqrt{1 - (x/A)^2} \] \[ a = -\omega^2 x \] At the mean position, x = 0, so acceleration (a) is zero and velocity is at maximum.
A particle is subjected to simple harmonic motions as: $ x_1 = \sqrt{7} \sin 5t \, \text{cm} $ $ x_2 = 2 \sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) \, \text{cm} $ where $ x $ is displacement and $ t $ is time in seconds. The maximum acceleration of the particle is $ x \times 10^{-2} \, \text{m/s}^2 $. The value of $ x $ is:
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?