A particle executes simple harmonic motion with an amplitude of 4 cm4 cm. At the mean position, the velocity of the particle is 10 cm s−110 cm s-1. The distance of the particle from the mean position when its speed becomes 5 cm s−15 cm s-1 is

A particle executes simple harmonicmotion with an amplitude of 5 cm.When the particle is at 4 cm from themean position, the magnitude of itsvelocity in SI units is equal to that of itsacceleration. Then, its periodic time inseconds is

A particle of mass 0.1 kg0.1 kg is executing simple harmonic motion of amplitude 0.1 m0.1 m. When the particle passes through the mean position, its kinetic energy is 8×10−3 J8×10-3 J. If the initial phase is 45°,45°, the equation of its motion is (Assume, x (t)𝑥 𝑡 as the position of the particle at time t𝑡)

A particle moves rectilinearly with a constant acceleration 1 m/s2. Its speed after 10 seconds is 5 m/s. The distance covered by the particle in this duration is (Initial & final velocities are in opposite direction)

particle of mass 250 g250 g executes a simple harmonic motion under a periodic force F=(–25x) N𝐹=(–25𝑥) N. The particle attains a maximum speed of 4 m s−14 m s-1 during its oscillation. The amplitude of the motion is ______cmcm

A particle in simple harmonic motion while passing through mean position will haveSelect one:a. Minimum kinetic energy and minimum potential energyb. Maximum kinetic energy and maximum potential energyc. Minimum kinetic energy and maximum potential energyd. Maximum kinetic energy and minimum potential energy