/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 A 0.150-kg toy is undergoing SHM... [FREE SOLUTION] | 91Ó°ÊÓ

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A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant \(k =\) 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s. What are the toy's (a) total energy at any point of its motion; (b) amplitude of motion; (c) maximum speed during its motion?

Short Answer

Expert verified
(a) Total energy is 0.0336 J. (b) Amplitude is 0.0133 m. (c) Maximum speed is 0.664 m/s.

Step by step solution

01

Find total energy

For a system in simple harmonic motion (SHM), the total mechanical energy is the sum of kinetic and potential energy. At any point, the total energy can be calculated as \[ E = rac{1}{2} k A^2 = rac{1}{2} k x^2 + rac{1}{2} m v^2 \]where \(x = 0.0120\, \text{m}\), \(v = 0.400\, \text{m/s}\), \(m = 0.150\, \text{kg}\), and \(k = 300\, \text{N/m}\). Substituting the known values:\[E = \frac{1}{2} (300) (0.0120)^2 + \frac{1}{2} (0.150) (0.400)^2 \E = 0.0216 + 0.012 = 0.0336\, \text{J}\]

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Key Concepts

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

Spring Constant
The spring constant, denoted by \( k \), is a measure that describes how stiff or strong a spring is. It is essential in determining how the spring will react to different forces. Specifically, it is part of Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position. This relationship is expressed as: \( F = -kx \), where:
  • \( F \) is the force exerted by the spring (in Newtons).
  • \( k \) is the spring constant (in N/m).
  • \( x \) is the displacement from the equilibrium position (in meters).
In the given exercise, the spring constant is 300 N/m. This indicates a relatively strong spring, capable of exerting a considerable force when compressed or stretched. Understanding the spring constant is critical for analyzing how the system behaves under SHM.
Total Mechanical Energy
The total mechanical energy of a system in Simple Harmonic Motion (SHM) is constant as long as no external forces (like friction) influence the motion. It comprises both kinetic energy (due to the motion of the mass) and potential energy (stored in the spring). Thus, the energy can be described by the equation:$$ E = \frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2} m v^2 $$
  • \( A \) is the amplitude, the maximum displacement from the equilibrium position.
  • \( x \) is the current displacement.
  • \( v \) is the velocity of the object.
  • \( m \) is the mass.
The exercise demonstrates this by calculating the total energy at a specified position. At 0.0120 m from equilibrium with a velocity of 0.400 m/s, the mechanical energy is 0.0336 J. This energy remains constant throughout the motion.
Amplitude of Motion
The amplitude of motion in SHM refers to the maximum distance the object moves from its equilibrium position. It is a crucial parameter because it dictates how far the object travels during each oscillation. The amplitude can be determined when the total mechanical energy is known, and the potential energy is at its maximum:- At the maximum displacement, the velocity is zero, and all the mechanical energy is potential energy.- The amplitude \( A \) can be found using the formula: \[ A = \sqrt{\frac{2E}{k}} \]Given the total energy calculated earlier as 0.0336 J and the spring constant \( k \) of 300 N/m, we can determine the amplitude of motion. This value signifies the greatest distance the system can reach from the central equilibrium position before returning.
Maximum Speed
In Simple Harmonic Motion, the maximum speed of the object occurs as it passes through the equilibrium position. At this point, all the mechanical energy is converted to kinetic energy, and thus the speed is at its highest. The maximum speed \( v_{max} \) can be calculated using the formula:\[ v_{max} = \sqrt{\frac{k}{m}} \cdot A \]
  • \( A \) is the amplitude of motion.
  • \( m \) is the mass of the object.
  • \( k \) is the spring constant.
The exercise requires finding this speed based on the given values. Determine amplitude \( A \) from the total energy and spring constant, and plug these values into the equation above to find \( v_{max} \). Recognizing and understanding where the velocity reaches its peak is key to mastering the dynamics of SHM.

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Most popular questions from this chapter

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.

An object with height \(h\), mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M\), and the cross- sectional area A of the object. You can ignore the damping due to fluid friction (see Section 14.7).

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the block is at \(x =\) 0.280 m, the acceleration of the block is \(-\)5.30 m/s\(^2\). What is the frequency of the motion?

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?

The tip of a tuning fork goes through 440 complete vibrations in 0.500 s. Find the angular frequency and the period of the motion.

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