The phase constant δ in SHM depends on the initial conditions (position and ve locity at t=0). Carefully consider the sign of the sine and cosine of δ to find the correct value
Using the given setup:
\[ \cos \theta = \frac{A}{2A} = \frac{1}{2} \]
From the trigonometric identity:
\[ \theta = \frac{\pi}{3} \]
The phase difference is given by:
\[ \delta = \frac{\pi}{2} - \frac{\pi}{3} \]
Simplify the expression:
\[ \delta = \frac{\pi}{6} \]
The values are:
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?

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: