The position of a harmonic oscillator at any time $t$ can be calculated using the equation:

$x(t) = A \cos(\omega t)$

where:

• $A$ is the amplitude of the motion (the initial displacement from equilibrium),
• $\omega$ is the angular frequency, defined as $\omega = \sqrt{\frac{k}{m}}$,
• $t$ is the time at which we want to find the position.

Given:

• $k = 6.24 \, \text{N/m}$,
• $m = 3.91 \, \text{kg}$,
• $A = 0.0267 \, \text{m}$,
• $t = 11.2 \, \text{s}$.

Step 1: Calculate Angular Frequency $\omega$

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{6.24}{3.91}} = \sqrt{1.5967} \approx 1.263 \, \text{rad/s}$

Step 2: Calculate Position $x(t)$ at $t = 11.2 \, \text{s}$

Using $x(t) = A \cos(\omega t)$:

$x(11.2) = 0.0267 \cos(1.263 \times 11.2)$

Calculating $1.263 \times 11.2$:

$1.263 \times 11.2 \approx 14.1456 \, \text{radians}$

Now, we find $\cos(14.1456)$:

$\cos(14.1456) \approx 0.736$

Thus:

$x(11.2) \approx 0.0267 \times 0.736 \approx 0.0196 \, \text{m}$

Answer

The position of the harmonic oscillator at $t = 11.2 \, \text{s}$ is approximately:

$x(11.2) = 0.0196 \, \text{m}$

Would you like a more detailed explanation or have any questions?

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1. What is the physical meaning of angular frequency in simple harmonic motion?
2. How would the position change if the mass of the oscillator were doubled?
3. What are the kinetic and potential energies at a given position?
4. How would changing the spring constant affect the oscillation frequency?
5. What is the maximum velocity of the oscillator, given the same initial conditions?

Tip: In simple harmonic motion, the phase of the cosine or sine function often determines the motion’s position relative to equilibrium at any time.