Two pendulums of length $121 \text{ cm}$ and $100 \text{ cm}$ start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x = rac{a}{2}$ will be:

The restoring force of a spring, with a block attached to the free end of the spring, is represented by:

The displacement-time ($x-t$) graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t = 2$ s is:

A particle executing simple harmonic motion has a kinetic energy of $K = K_0 \cos^2(\omega t)$. The values of the maximum potential energy and the total energy are, respectively:

When two displacements represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ are superimposed, the motion is:

From the given functions, identify the function which represents a periodic motion:

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in seconds is:

Match List-I with List-II. **List-I (Graphs)** (a) Displacement-time graph with decreasing amplitude (Damped) (b) Displacement-time graph with constant amplitude (Undamped) (c) Force-displacement graph (Linear relationship, F = -kx) (d) Energy-time graph (Constant total energy) **List-II (Situations)** (i) Total mechanical energy is conserved (ii) Bob of a pendulum is oscillating under negligible air friction (iii) Restoring force of a spring (iv) Bob of a pendulum is oscillating along with air friction Choose the correct answer from the options given below: