A harmonic oscillator is described by the function x(t) = (0.500 m) cos(0.290t). Find the oscillator’s maximum velocity and maximum acceleration. Find the oscillator’s position, velocity, and acceleration when t = 1.50 s.
The correct answer and explanation is:
To solve this harmonic oscillator problem, we begin with the given displacement function: x(t)=(0.500 m)cos(0.290t)x(t) = (0.500\, \text{m}) \cos(0.290t)
This function represents simple harmonic motion (SHM) of the form: x(t)=Acos(ωt)x(t) = A \cos(\omega t)
Where:
- A=0.500 mA = 0.500 \, \text{m} is the amplitude,
- ω=0.290 rad/s\omega = 0.290 \, \text{rad/s} is the angular frequency.
🔹 Maximum Velocity and Acceleration
- Maximum velocity is given by:
vmax=Aω=(0.500)(0.290)=0.145 m/sv_{\text{max}} = A\omega = (0.500)(0.290) = 0.145 \, \text{m/s}
- Maximum acceleration is given by:
amax=Aω2=(0.500)(0.290)2=0.500×0.0841=0.04205 m/s2a_{\text{max}} = A\omega^2 = (0.500)(0.290)^2 = 0.500 \times 0.0841 = 0.04205 \, \text{m/s}^2
🔹 At Time t=1.50 st = 1.50\, \text{s}
We differentiate x(t)x(t) to find velocity and acceleration:
- Position:
x(1.50)=0.500cos(0.290×1.50)=0.500cos(0.435)≈0.500×0.906=0.453 mx(1.50) = 0.500 \cos(0.290 \times 1.50) = 0.500 \cos(0.435) \approx 0.500 \times 0.906 = 0.453\, \text{m}
- Velocity (first derivative):
v(t)=dxdt=−Aωsin(ωt)v(t) = \frac{dx}{dt} = -A\omega \sin(\omega t) v(1.50)=−0.500×0.290×sin(0.435)≈−0.145×0.421=−0.061 m/sv(1.50) = -0.500 \times 0.290 \times \sin(0.435) \approx -0.145 \times 0.421 = -0.061\, \text{m/s}
- Acceleration (second derivative):
a(t)=d2xdt2=−Aω2cos(ωt)a(t) = \frac{d^2x}{dt^2} = -A\omega^2 \cos(\omega t) a(1.50)=−0.500×(0.290)2×cos(0.435)=−0.04205×0.906=−0.0381 m/s2a(1.50) = -0.500 \times (0.290)^2 \times \cos(0.435) = -0.04205 \times 0.906 = -0.0381\, \text{m/s}^2
✅ Final Answers:
- Maximum velocity: 0.145 m/s0.145 \, \text{m/s}
- Maximum acceleration: 0.04205 m/s20.04205 \, \text{m/s}^2
- At t=1.50 st = 1.50\, \text{s}:
- Position: x=0.453 mx = 0.453 \, \text{m}
- Velocity: v=−0.061 m/sv = -0.061 \, \text{m/s}
- Acceleration: a=−0.0381 m/s2a = -0.0381 \, \text{m/s}^2
🧠 Explanation (Conceptual)
In SHM, the object oscillates back and forth around an equilibrium point. The cosine function implies it starts at maximum displacement. Velocity is greatest as the mass passes through equilibrium, and acceleration is greatest at maximum displacement. These values depend on both amplitude and angular frequency. By applying calculus to the position function, we obtained velocity and acceleration at any time, such as t=1.50 st = 1.50\, \text{s}.