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
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,
How to determine maximum amplitude of oscillation of a system, using the question below? A large block P executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency f=1.50Hz. Block B rest
Material point is accomplishing harmonic oscillation. What is the time for its displacement from equilibrium position to half of its amplitude ? The period of oscillation is 24 s, and the initial phase is 0 rad.
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
For a damped simple harmonic oscillator, the block has a mass of 2.4 kg and the spring constant is 14 N/m. The damping force is given by -b(dx/dt), where b = 260 g/s. The block is pulled down 11.0 cm and released. (a) Calculate
When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12 Hz. When another object of mass m2 is hung on the spring along with m1, the frequency of the motion is 4 Hz.