Ch 14: Periodic Motion

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 14: Periodic Motion

Problem 8

Chapter 14, Problem 8

In a physics lab, you attach a 0.200-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.

Verified step by step guidance

First, understand that the problem involves simple harmonic motion (SHM) of a glider attached to a spring. The time given is the period of half an oscillation, so the full period (T) is twice this time: T = 2 * 2.60 s.

Recall the formula for the period of a mass-spring system in SHM:

T = \frac{2 π}{\sqrt{\frac{m}{k}}}

, where T is the period, m is the mass, and k is the spring constant.

Rearrange the formula to solve for the spring constant k:

k = \frac{m}{\frac{4 π^2}{T^{2}}}

Substitute the known values into the equation: m = 0.200 kg and T = 5.20 s (since T is twice the given time).

Calculate the spring constant k using the rearranged formula and the substituted values. This will give you the spring's force constant.

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Key Concepts

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

Simple Harmonic Motion

Simple Harmonic Motion (SHM) describes the oscillatory motion of systems like springs and pendulums, where the restoring force is proportional to the displacement from equilibrium. In this problem, the glider's motion through the equilibrium point indicates SHM, characterized by sinusoidal oscillations and a constant period.

Period of Oscillation

The period of oscillation is the time taken for one complete cycle of motion in SHM. It is crucial for determining the properties of the system, such as the spring constant. In this scenario, the time from the first to the second equilibrium point passage is half the period, helping calculate the full period and subsequently the spring constant.

Spring Constant

The spring constant, denoted as k, measures the stiffness of a spring and is defined by Hooke's Law, F = -kx, where F is the restoring force and x is the displacement. It can be derived from the period of oscillation using the formula T = 2π√(m/k), where T is the period and m is the mass of the glider.

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Related Practice

Textbook Question

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.70 s to travel from x = 0.090 m to x = -0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel from x = 0.180 m to x = -0.180 m?

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Textbook Question

The displacement of an oscillating object as a function of time is shown in Fig. E14.4. What is (c) the period? (d) the angular frequency of this motion?

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Textbook Question

A 2.40-kg ball is attached to an unknown spring and allowed to oscillate. Figure E14.7 shows a graph of the ball's position x as a function of time t. What are the oscillation's (a) period, (b) frequency, (c) angular frequency, and (d) amplitude? (e) What is the force constant of the spring?

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Textbook Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

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Textbook Question

A machine part is undergoing SHM with a frequency of 4.00 Hz and amplitude 1.80 cm. How long does it take the part to go from x = 0 to x = -1.80 cm ?

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Textbook Question

An object is undergoing 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. (b) from x = 0.160 m to x = 0.

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