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

To determine the position of the oscillation at t = 2s, we need to use the equation of motion for a simple harmonic oscillator:

x(t) = A * cos(ωt + φ)

Where:
- x(t) is the position of the oscillation at time t,
- A is the amplitude of the oscillation,
- ω is the angular frequency, given by ω = sqrt(k / m),
- t is the time,
- φ is the phase constant.

Given:
- m = 0.5 kg (mass of the oscillator),
- k = 10 N/m (spring constant),
- A = 3 cm = 0.03 m (amplitude),
- t = 2 s (time).

First, we can find the angular frequency using the given spring constant and mass:

ω = sqrt(k / m)
= sqrt(10 N/m / 0.5 kg)
= sqrt(20) rad/s
≈ 4.47 rad/s

Next, we can calculate the position at t = 2s by substituting the values into the equation of motion:

x(2) = A * cos(ω * t + φ)

Since the oscillation is released from rest at x = A when t = 0, we know that the initial phase φ is 0.

x(2) = A * cos(ω * 2 + φ)
= 0.03 m * cos(4.47 rad/s * 2 + 0)
≈ 0.03 m * cos(8.94 rad)
≈ 0.03 m * (-0.808)
≈ -0.024 m

Therefore, the position of the oscillation at t = 2s is approximately -0.024 meters.