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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 \(x=-0.180 \mathrm{m}\) .

Short Answer

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

Step by step solution

01

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 \( \omega = \frac{2\pi}{1.500} \approx 4.189 \text{ rad/s} \).
02

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:\[ x(0.5) = 0.240 \cos(4.189 \cdot 0.5) \approx 0.240 \cos(2.094) = 0.240 \times (-0.5) = -0.120 \text{ m} \]
03

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 \( x(0.5) = -0.120 \text{ m} \):\[ F = -0.351 \times (-0.120) \approx 0.042 \text{ N (in the positive x direction)} \]
04

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:\[ t \approx \frac{2.418}{4.189} \approx 0.577 \text{ s} \]
05

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:\[ v = 4.189 \sqrt{0.240^2 - (-0.180)^2} \approx 4.189 \sqrt{0.0576 - 0.0324} \approx 4.189 \sqrt{0.0252} \approx 4.189 \times 0.1588 \approx 0.665 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

CP SHM of a Butcher's Scale. A spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) is hung vertically, and a 0.200 -kg pan is suspended from its lower end. A butcher drops a 2.2 -kg steak onto the pan from a height of 0.40 \(\mathrm{m}\) . The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

In a physics lab, you attach a \(0.200-\mathrm{kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

CALC A Spring with Mass. The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring's mass, consider a spring with mass \(M\) , equilibrium length \(L_{0}\), and spring constant \(k\). When stretched or compressed to a length \(L,\) the potential energy is \(\frac{1}{2} k x^{2},\) where \(x=L-L_{0-}\) (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\) . Assume that the speed of points along the length of the spring varies linearly with distance \(I\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L\) . The result is not \(\frac{1}{2} M v^{2}\) , since not all of the spring moves with the same speed.) (b) Take the time derivative of the conservation of energy equation, Eq. (14.21), for a mass \(m\) moving on the end of a massless. By comparing your results to Eq. (14.8), which defines \(\omega,\) show that the angular frequency of oscillation is \(\omega=\sqrt{k / m}\) (c) Apply the procedure of part (b) to obtain the angular frequency of oscillation \(\omega\) of the spring considered in part (a). If the effective mass \(M^{\prime}\) of the spring is defined by \(\omega=\sqrt{k / M^{\prime}},\) what is \(M^{\prime}\) in terms of \(M ?\)

A 1.80 -kg connecting rod from a car engine is pivoted about a horizontal knife edge as shown in Fig. E14.53. The center of gravity of the rod was located by balancing and is 0.200 \(\mathrm{m}\) from the pivot. When the rod is set into small-amplitude oscillation, it makes 100 complete swings in 120 s. Calculate the moment of inertia of the rod about the rotation axis through the pivot.

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