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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 uniform, solid metal disk of mass 6.50 kg and diameter 24.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.23 N tangent to the rim of the disk to turn it by 3.34\(^\circ\), thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.

A 1.35-kg object is attached to a horizontal spring of force constant 2.5 N/cm. The object is started oscillating by pulling it 6.0 cm from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.5 cm. (a) How much energy has this system lost to damping during these eight cycles? (b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

An object is undergoing SHM with period 1.200 s and amplitude 0.600 m. At \(t =\) 0 the object is at \(x =\) 0 and is moving in the negative \(x\)-direction. How far is the object from the equilibrium position when \(t =\) 0.480 s?

Two pendulums have the same dimensions (length \(L\)) and total mass ( \(m\) ). Pendulum \(A\) is a very small ball swinging at the end of a uniform massless bar. In pendulum \(B\), half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

A large, 34.0-kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is 0.60 m below the pivot. The bell's moment of inertia about an axis at the pivot is 18.0 kg \(\cdot\) m\(^2\). The clapper is a small, 1.8-kg mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently\(-\)that is, for the period of oscillation for the bell to equal that of the clapper?

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