Physics
posted by Anonymous
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, moving away from equilibrium?

s
tf
Respond to this Question
Similar Questions

conceptual physics problem
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 amplitudes … 
Physics (Oscillation)
If a particle moves in simple harmonic motion with a frequency of 3.00 Hz and an amplitude of 5.00 cm through what total distance does the particle move during one cycle of its motion, what is its maximum speed, where does this maximum … 
physics
consider a simple harmonic oscillation with m=0.5kg,k=10N/m and amplitude A=3cm,if the oscillation is released from rest at x=A when the clock is set to t=0,determine the position of the oscillation at t=2s 
physics
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 … 
Physics
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 particle … 
Physics: Simple Harmonic Motion
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 particle … 
chem
How to determine maximum amplitude of oscillation of a system, using the question below? 
physics
for a damped harmonic oscillation , the equation of motion is md^2x/dt^2+(gamma)dx/dt+kx=o with m=0.025kg,(gamma)=0.07kg/s and k=85N/m. *calculate the period of motion *number of oscillation in which its amplitude will become half … 
physics
A particle that undergoes simple harmonic motion has a period of 0.4 s and an amplitude of 12 mm. The maximam velocity of the particle is? 
Physics
A particle executing simple harmonic motion of amplitude 5 cm and period 2 s. Find the speed of the particle at a point where its acceleration is half of its maximum value