Ch. 14 - Oscillations

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 14 - Oscillations

Problem 35a

Chapter 14, Problem 35a

A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end of the spring is fixed to a wall. It takes 3.2 J of work to compress the spring by 0.13 m. The mass is then released from rest and experiences a maximum acceleration of 12m/s². Find the value of the spring constant.

Verified step by step guidance

Step 1: Recall the formula for the work done on a spring, which is given by \( W = \frac{1}{2} k x^2 \), where \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the compression or extension of the spring. Here, \( W = 3.2 \; \text{J} \) and \( x = 0.13 \; \text{m} \). Rearrange the formula to solve for \( k \): \( k = \frac{2W}{x^2} \).

Step 2: Substitute the given values into the formula. Use \( W = 3.2 \; \text{J} \) and \( x = 0.13 \; \text{m} \). The equation becomes \( k = \frac{2(3.2)}{(0.13)^2} \).

Step 3: Simplify the denominator \( (0.13)^2 \) and the numerator \( 2(3.2) \). Then divide the numerator by the denominator to find the spring constant \( k \).

Step 4: Verify the units of \( k \). Since work \( W \) is in joules (\( \text{kg} \cdot \text{m}^2 / \text{s}^2 \)) and \( x \) is in meters, the units of \( k \) will be \( \text{N/m} \), which is consistent with the spring constant's unit.

Step 5: Conclude that the spring constant \( k \) has been determined using the given data and the work-energy relationship for a spring.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hooke's Law

Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position, expressed as F = -kx, where F is the force, k is the spring constant, and x is the displacement. This principle is fundamental in understanding how springs behave under compression or extension.

Work-Energy Principle

The Work-Energy Principle states that the work done on an object is equal to the change in its kinetic energy. In the context of the spring, the work done to compress it is stored as potential energy, which is converted into kinetic energy when the mass is released, allowing us to relate work, energy, and motion.

Acceleration and Newton's Second Law

Newton's Second Law relates the acceleration of an object to the net force acting on it and its mass, expressed as F = ma. In this scenario, the maximum acceleration experienced by the mass can be used to determine the net force exerted by the spring, which is crucial for calculating the spring constant using the relationship between force, displacement, and the spring's properties.

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Related Practice

Textbook Question

Agent Arlene devised the following method of measuring the muzzle velocity of a rifle (Fig. 14–34). She fires a bullet into a 4.148-kg wooden block resting on a smooth surface, and attached to a spring of spring constant k = 162.7 N/m. The bullet, whose mass is 7.450 g, remains embedded in the wooden block. She measures the maximum distance that the block compresses the spring to be 9.460 cm. What is the speed υ of the bullet?

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Determine the phase constant ϕ in Eq. 14–4 if, at t = 0, the oscillating mass is at 𝓍 = A .

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At t = 0, an 885-g mass at rest on the end of a horizontal spring (k = 184 N/m) is struck by a hammer which gives it an initial speed of 2.12 m/s. Determine the period and frequency of the motion.

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A 0.25-kg mass at the end of a spring oscillates 3.2 times per second with an amplitude of 0.15 m. Determine the equation describing the motion of the mass, assuming that at t = 0, 𝓍 was a maximum.

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Textbook Question

An object with mass 2.7 kg is executing simple harmonic motion, attached to a spring with spring constant k = 310 N/m. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.60 m/s. Calculate the maximum speed attained by the object.

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