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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 m/s to the right and an acceleration of 8.40 m/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?

Short Answer

Expert verified
0.240 m

Step by step solution

01

Identify Given Values

We start by identifying the given details of the problem. The object is in Simple Harmonic Motion (SHM):- Displacement from equilibrium, \( x = 0.600 \ \text{m} \) (to the right),- Velocity, \( v = 2.20 \ \text{m/s} \) (to the right),- Acceleration, \( a = -8.40 \ \text{m/s}^2 \) (to the left).
02

Use Acceleration Formula in SHM

In SHM, acceleration is given by \( a = - \omega^2 x \). Rearranging to solve for angular frequency \( \omega \), we have:\[a = -\omega^2 x \8.40 = \omega^2 \times 0.600 \\omega^2 = \frac{8.40}{0.600} = 14 \\omega = \sqrt{14}\]
03

Calculate Maximum Displacement

In SHM, when the object stops momentarily, it is at its maximum displacement \( x_{max} \). Using energy conservation, total energy at maximum displacement is equal to total energy at \( x = 0.600 \ \text{m} \):\[\frac{1}{2}m\omega^2 x_{max}^2 = \frac{1}{2}m\omega^2 x^2 + \frac{1}{2}mv^2 \\omega^2 x_{max}^2 = \omega^2 x^2 + v^2 \x_{max}^2 = x^2 + \frac{v^2}{\omega^2} \x_{max}^2 = (0.600)^2 + \frac{(2.20)^2}{14} \x_{max}^2 = 0.360 + 0.346 \x_{max} = \sqrt{0.706} \]
04

Solve for Additional Displacement

Now calculate the additional distance from the given point:\[ \text{Maximum displacement: } x_{max} = \sqrt{0.706} \approx 0.840 \ \text{m} \\text{Initial displacement: } x = 0.600 \ \text{m} \\text{Additional distance = } x_{max} - x = 0.840 - 0.600 \]
05

Final Calculation

Perform the subtraction to find the additional displacement:\[\text{Additional displacement } = 0.240 \ \text{m}\]

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Key Concepts

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

Angular Frequency
In Simple Harmonic Motion (SHM), angular frequency is a crucial concept. It is represented by the symbol \( \omega \) and serves as a measure of how quickly a system oscillates. Angular frequency can be derived from the formula \( a = - \omega^2 x \), where \( a \) is the acceleration at displacement \( x \). This formula shows the relationship of acceleration to angular frequency and displacement.For example, in our problem, at a displacement of 0.600 meters, the object has an acceleration of -8.40 m/s\(^2\). Solving \( 8.40 = \omega^2 \times 0.600 \) gives \( \omega^2 = 14 \), leading to \( \omega = \sqrt{14} \). Understanding this concept allows us to predict how fast the oscillations happen and to link this to energy and displacement in SHM.
Maximum Displacement
Maximum displacement in SHM refers to the furthest point the object reaches from its equilibrium position before momentarily coming to rest. This point marks the peak of its oscillation where velocity is zero.We use energy conservation principles to calculate this. Both potential and kinetic energies are in play, but at maximum displacement, all energy is potential. The formula to find maximum displacement from the point at \( x = 0.600 \, \text{m} \) is given by:\[ \omega^2 x_{max}^2 = \omega^2 x^2 + v^2 \] Solving gives \[ x_{max}^2 = 0.360 + 0.346 \quad \Rightarrow \quad x_{max} = \sqrt{0.706} \approx 0.840 \, \text{m} \] This shows the farthest point the object reaches before reversing direction.
Energy Conservation
Energy conservation in SHM is essential to understanding how the system oscillates. The total mechanical energy remains constant; it simply shifts between kinetic and potential forms.At any point, the energy \( E \) is\[ E = \frac{1}{2} m \omega^2 x^2 + \frac{1}{2} mv^2 \] At maximum displacement, kinetic energy is zero, and all energy is potential:\[ E = \frac{1}{2} m \omega^2 x_{max}^2 \] The conservation principle lets us equate these energies, allowing us to solve for variables like \( x_{max} \). This principle helps explain how the changing energy forms maintain the harmonic motion.
Velocity
Velocity plays a dynamic role in SHM as it varies continuously with displacement. While displacement and acceleration at any point can determine velocity, in SHM, these components oscillate in sinusoidal patterns.From its maximum when passing through equilibrium to zero at maximum displacement, velocity is crucial for analyzing motion. Given:\[ v = 2.20 \, \text{m/s at} \, x = 0.600 \, \text{m} \] This velocity helps us set the energy equations by providing kinetic energy, which partners with potential energy to maintain the motion. Observing these oscillations reveals insights into the system's dynamics. Understanding this shows how all components interrelate in harmonic motion.

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Most popular questions from this chapter

A spring of negligible mass and force constant \(k =\) 400 N/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 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?

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

While on a visit to Minnesota ("Land of 10,000 Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the 1500-kg boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 cm. The boat takes 3.5 s to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 kg) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 cm of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

A 0.500-kg glider, attached to the end of an ideal spring with force constant \(k =\) 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x = -\)0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x = -\)0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

An object is undergoing \(\textbf{SHM}\) with period 0.900 s and amplitude 0.320 m. At \(t =\) 0 the object is at \(x =\) 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x =\) 0.320 m to \(x =\) 0.160 m and (b) from \(x =\) 0.160 m to \(x =\) 0.

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