We can model this problem using the equation for simple harmonic motion (SHM), which arises in spring-mass systems governed by Hooke’s law. The position of the mass as a function of time is described by the equation:

$x(t) = A \cos(\omega t) + B \sin(\omega t)$

Step 1: Hooke’s Law and Determining Spring Constant $k$

We start by using Hooke's Law, which states: $F = kx$ The restoring force $F$ due to the spring is equal to the weight of the mass: $F = mg$ where:

• $m = 3$ grams $= 0.003$ kg (since 1 gram = 0.001 kg),
• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $x = 4 \, \text{cm} = 0.04 \, \text{m}$ (the spring’s stretch due to the mass).

From Hooke’s law: $mg = kx$ Solving for $k$: $k = \frac{mg}{x} = \frac{0.003 \times 9.8}{0.04} = 0.735 \, \text{N/m}$

Step 2: Determining Angular Frequency $\omega$

The angular frequency $\omega$ of SHM is given by: $\omega = \sqrt{\frac{k}{m}}$ Substitute $k = 0.735 \, \text{N/m}$ and $m = 0.003 \, \text{kg}$: $\omega = \sqrt{\frac{0.735}{0.003}} = \sqrt{245} \approx 15.652 \, \text{rad/s}$

Step 3: Initial Conditions

The initial conditions given in the problem are:

1. The initial displacement $x(0) = -0.04 \, \text{m}$ (since the spring stretches downward by 4 cm),
2. The initial velocity $v(0) = 6 \, \text{cm/s} = 0.06 \, \text{m/s}$ (upward).

Using the general solution $x(t) = A \cos(\omega t) + B \sin(\omega t)$, we apply these initial conditions.

Initial Displacement:

At $t = 0$, $x(0) = A \cos(0) + B \sin(0)$, so: $x(0) = A = -0.04 \, \text{m}$ Thus, $A = -0.04$.

Initial Velocity:

The velocity is the time derivative of the position: $v(t) = \frac{dx(t)}{dt} = -A \omega \sin(\omega t) + B \omega \cos(\omega t)$ At $t = 0$, $v(0) = 0.06 \, \text{m/s}$, so: $v(0) = B \omega = 0.06$ Substituting $\omega = 15.652$: $B \times 15.652 = 0.06 \quad \Rightarrow \quad B = \frac{0.06}{15.652} \approx 0.00383$

Step 4: Position Function

The equation of motion is: $x(t) = -0.04 \cos(15.652 t) + 0.00383 \sin(15.652 t)$

Step 5: Determining Amplitude, Period, and Frequency

• Amplitude: The amplitude $A_{\text{total}}$ is given by: $A_{\text{total}} = \sqrt{A^2 + B^2} = \sqrt{(-0.04)^2 + (0.00383)^2} \approx 0.04018 \, \text{m} = 4.018 \, \text{cm}$

• Period: The period $T$ of the motion is: $T = \frac{2\pi}{\omega} = \frac{2\pi}{15.652} \approx 0.4015 \, \text{seconds}$

• Frequency: The frequency $f$ is the reciprocal of the period: $f = \frac{1}{T} = \frac{1}{0.4015} \approx 2.49 \, \text{Hz}$

Final Answer

The position of the mass at any given time $t$ is: $x(t) = -0.04 \cos(15.652 t) + 0.00383 \sin(15.652 t)$

• Amplitude: $4.018 \, \text{cm}$
• Period: $0.4015 \, \text{seconds}$
• Frequency: $2.49 \, \text{Hz}$

Would you like more details on any part of this solution?

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1. What is the effect of damping on this system?
2. How does the mass affect the angular frequency?
3. What happens if we change the spring constant?
4. How can the amplitude be increased without changing mass or spring?
5. Why is initial velocity upward instead of downward?

Tip: Angular frequency $\omega$ directly depends on both the spring constant and the mass, so increasing $k$ or decreasing $m$ increases $\omega$.