A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t=0 it is at position x=5 cm going towards positive x direction . Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t=4s.
A particle executes simple harmonic motion such that at t = 0 it is at the amplitude of oscillation A = 22.5 cm. The period of the oscillation is 0.25 s. When is the first time this particle will be at x = -1/2 the amplitude,
A particle is oscillating in simple harmonic motion. The time required for the particle to travel through one complete cycle is equal to the period of the motion, no matter what the amplitude is. But how can this be, since larger
A 10kg particle undergoes simple harmonic motion with an amplitude of 2.0mm, a maximum acceleration of 8.0x10^3 m/s^2, and an unknown phase constant (phi) What are: a.) the period of the motion b.) the maximum speed of the
A simple harmonic oscillator executes motion whose amplitude is 0.20 m and it completes 60 oscillations in 2 minutes. i) Calculate its time period and angular frequency. ii) If the initial phase is 45°, write expressions for
A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the system is 2.00 J, find (a) The force constant of the spring and (b) The amplitude of the