A 170g mass attached to a horizontal spring oscillates at a frequency of 3.10Hz . At t =0s, the mass is at x= 6.60cm and has vx = − 15.0cm/s. Determine the position at t= 5.00s.
The period is 0.32258 s
The angular frequency is 19.478 rad/s
The amplitude is 6.6448×10^−2 m
The phase constant is 0.1162 rad
The maximum speed is 1.29 m/s
The maximum acceleration 25.2 m/s^2
The total energy is 0.142 J
The position at t= 5.00 s is….????
To determine the position at t=5.00s, we need to use the equation for simple harmonic motion:
x(t) = A * cos(ω * t + φ),
where x(t) represents the position at time t, A is the amplitude of the motion, ω is the angular frequency, t is time, and φ is the phase constant.
Given:
Amplitude (A) = 6.6448×10^−2 m
Angular frequency (ω) = 19.478 rad/s
Phase constant (φ) = 0.1162 rad
Time (t) = 5.00s
We can now substitute the given values into the equation to find the position at t=5.00s:
x(t) = (6.6448×10^−2 m) * cos((19.478 rad/s) * (5.00s) + 0.1162 rad)
Calculating the equation:
x(t) = (6.6448×10^−2 m) * cos(97.39 rad + 0.1162 rad)
x(t) = (6.6448×10^−2 m) * cos(97.5062 rad)
Now, calculate the cosine of 97.5062 rad:
cos(97.5062 rad) ≈ 0.0645
Substituting this value back into the equation:
x(t) ≈ (6.6448×10^−2 m) * 0.0645
Calculating the final value:
x(t) ≈ 4.29 × 10^−3 m
Therefore, the position at t=5.00s is approximately 4.29 × 10^−3 m.