Problem 84 - University Physics with Modern Physics

Question

A 0.0200 -kg bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;\) (c) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;\) (d) the speed of the bolt when \(x=-0.180 \mathrm{m}\) .

Step-by-Step Solution

Answer

(a) -0.120 m, (b) 0.042 N (positive x), (c) 0.577 s, (d) 0.665 m/s

Step 1: Determine the Angular Frequency

The angular frequency \( \omega \) can be calculated using the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period. For a period of 1.500 s, we find \( \omega = \frac{2\pi}{1.500} \approx 4.189 \text{ rad/s} \).

Step 2: Find the Displacement at t=0.5s

Using the simple harmonic motion equation \( x(t) = A \cos(\omega t + \phi) \), where \( A = 0.240 \text{ m} \) and the initial phase \( \phi = 0 \text{ rad} \) since the maximum amplitude is at \( t=0 \). Substitute \( t=0.5 \text{ s} \) and calculate:\[ x(0.5) = 0.240 \cos(4.189 \cdot 0.5) \approx 0.240 \cos(2.094) = 0.240 \times (-0.5) = -0.120 \text{ m} \]

Step 3: Calculate the Force at t=0.5s

The force acting on the bolt can be determined using Hooke's Law for SHM, \( F = -kx \). First, find the spring constant \( k \) using \( \omega^2 = \frac{k}{m} \) which gives \( k = m \omega^2 = 0.02 \times (4.189)^2 \approx 0.351 \text{ N/m} \). Then, use \( x(0.5) = -0.120 \text{ m} \):\[ F = -0.351 \times (-0.120) \approx 0.042 \text{ N (in the positive x direction)} \]

Step 4: Determine the Minimum Time to Reach x=-0.180m

Using the equation \( x(t) = A \cos(\omega t + \phi) \), solve for \( t \) when \( x = -0.180 \text{ m} \). This requires solving \( -0.180 = 0.240 \cos(4.189t) \), giving \( \cos(4.189t) = -0.75 \). The arccosine gives multiple solutions, but the first positive solution is \( 4.189t = \cos^{-1}(-0.75) \approx 2.418 \) which implies:\[ t \approx \frac{2.418}{4.189} \approx 0.577 \text{ s} \]

Step 5: Calculate the Speed at x=-0.180m

The speed in SHM can be calculated using \( v = \omega \sqrt{A^2 - x^2} \). For \( x = -0.180 \text{ m} \), the speed is:\[ v = 4.189 \sqrt{0.240^2 - (-0.180)^2} \approx 4.189 \sqrt{0.0576 - 0.0324} \approx 4.189 \sqrt{0.0252} \approx 4.189 \times 0.1588 \approx 0.665 \text{ m/s} \]

Key Concepts

Angular Frequency

Angular frequency is a crucial concept when studying simple harmonic motion (SHM). It helps us understand how rapidly the oscillations occur in a system. Angular frequency, denoted by \( \omega \), is calculated by the formula: \[ \omega = \frac{2\pi}{T} \] Here, \( T \) represents the period, which is the time taken for one complete cycle of oscillation. The angular frequency gives a sense of how many oscillations occur per unit time but in terms of radians per second, as it's connected to the circle. Typically:

• Higher \( \omega \) means the system oscillates more rapidly.
• Lower \( \omega \) means the system oscillates more slowly.

In our example, with a period of 1.500 seconds, the angular frequency was calculated to be approximately \( 4.189 \text{ rad/s} \). This tells us the bolt completes its oscillation fairly quickly, indicating a brisk harmonic motion.

Displacement Calculation

Displacement in simple harmonic motion describes how far an object has moved from its equilibrium or central position at a given time. This is derived using:\[ x(t) = A \cos(\omega t + \phi) \]where:

• \( A \) is the amplitude, the maximum distance from the equilibrium.
• \( \omega \) is the angular frequency.
• \( t \) is the time at which you want to find the displacement.
• \( \phi \) is the phase angle; it adjusts the starting point of the wave. In our case, it is 0 because oscillation starts at its peak.

This equation computes how the position of the object changes over time. Using this at \( t = 0.5 \text{ s} \), we can calculate the displacement to be \( -0.120 \text{ m} \), indicating that the bolt has moved 0.120 meters in the opposite direction from its beginning point at maximum amplitude.

Hooke's Law

Hooke's Law is a fundamental principle governing the forces in systems where spring-like restoring forces dominate. It states:\[ F = -kx \]In this formula:

• \( F \) is the force exerted.
• \( k \) is the spring constant, representing the stiffness of the spring.
• \( x \) is the displacement from the equilibrium position.

It's essential in understanding how the force varies with displacement. Often, the negative sign indicates the force is always directed towards the equilibrium position (opposite to the displacement). From our problem, with a displacement \( x = -0.120 \text{ m} \), the force calculated was approximately \( 0.042 \text{ N} \), showing the force direction is opposite to the displacement indicating a restoring force towards equilibrium.

Spring Constant

The spring constant \( k \) defines how stiff a spring is in simple harmonic motion. A larger \( k \) means the spring (or spring-like mechanism) is stiffer and requires more force to displace it by a certain amount. It can be derived using:\[ \omega^2 = \frac{k}{m} \]where:

• \( \omega \) is the angular frequency.
• \( m \) is the mass of the object.
• \( k \) is what we're solving for, the spring constant.

In the exercise, substituting the given values gave us \( k \approx 0.351 \text{ N/m} \). This value helps in predicting how much force will be necessary for a given displacement. The spring constant is key in calculating the system's dynamics and its future motion behavior, such as displacements and frequencies.