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A \(55.0 \mathrm{~g}\) block oscillates in SHM on the end of a spring with \(k=1500 \mathrm{~N} / \mathrm{m}\) according to \(x=x_{m} \cos (\omega t+\phi) .\) How long does the block take to move from position \(+0.800 x_{m}\) to (a) position \(+0.600 x_{m}\) and \((\mathrm{b})\) position \(-0.800 x_{m} ?\)

Short Answer

Expert verified
(a) 0.0017 s; (b) 0.0285 s.

Step by step solution

01

Identify the Known Values

We are given the mass of the block as \(55.0\, \mathrm{g} = 0.055\, \mathrm{kg}\) and the spring constant \(k = 1500\, \mathrm{N/m}\). The positions \(x_{1} = 0.800x_{m}\), \(x_{2} = 0.600x_{m}\), and \(x_{3} = -0.800x_{m}\) are given in terms of maximum displacement \(x_{m}\).
02

Determine the Angular Frequency

The angular frequency \(\omega\) is calculated using the formula for simple harmonic motion: \(\omega = \sqrt{\frac{k}{m}}\). Substituting the available values: \(\omega = \sqrt{\frac{1500}{0.055}} = 165.2\, \mathrm{rad/s}\).
03

Set Up the Equation for Motion

Using the equation \(x = x_{m} \cos(\omega t + \phi)\), we make use of the positions and solve for time differences. For position \(x = 0.800x_{m}\), \(0.800 = \cos(\omega t_{1} + \phi)\) leading to \(\omega t_{1} + \phi = \cos^{-1}(0.800)\). For \(x = 0.600x_{m}\), \(0.600 = \cos(\omega t_{2} + \phi)\).
04

Solve for Time (a)

Find the time difference from \(x = 0.800x_{m}\) to \(x = 0.600x_{m}\). Using \(\omega t_{1} + \phi = \cos^{-1}(0.800)\) and \(\omega t_{2} + \phi = \cos^{-1}(0.600)\), calculate \(\Delta t = t_{2} - t_{1}\ = \frac{\cos^{-1}(0.600) - \cos^{-1}(0.800)}{\omega} = \frac{0.927 - 0.644}{165.2} \approx 0.0017\, \mathrm{s}\).
05

Solve for Time (b)

For movement from \(x = 0.800x_{m}\) to \(x = -0.800x_{m}\), consider \(\omega t_{3} + \phi = \pi - \cos^{-1}(0.800)\) since \(\cos^{-1}(-0.800) = \pi - \cos^{-1}(0.800)\). Thus, \(\Delta t = t_{3} - t_{1} = \frac{(\pi - \cos^{-1}(0.800)) - \cos^{-1}(0.800)}{\omega} = \frac{\pi - 2\times 0.644}{165.2} \approx 0.0285\, \mathrm{s}\).

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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 as \(k\), is a crucial parameter in simple harmonic motion. It describes the stiffness of a spring.
A high spring constant means the spring is very stiff and requires more force to stretch it, while a lower constant means the spring is more easily deformed.
The SI unit for spring constant is Newtons per meter (N/m). In the context of simple harmonic motion (SHM), the spring constant determines how "tight" the spring’s force is, directly affecting the dynamics of any attached block or mass.
The force that the spring exerts can be expressed using Hooke's Law:
  • Hooke's Law: \( F = -kx \)
Here, \(F\) is the force exerted by the spring, \(x\) is the displacement from the equilibrium position, and \(-k\) indicates that the force is always directed opposite to the displacement. This relationship ensures that the spring complies with SHM's conditions, promoting periodic motion.
Angular Frequency
Angular frequency \(\omega\) is a measure of how quickly an object oscillates back and forth in simple harmonic motion.
It can be thought of as the rate of rotation in the SHM system, giving an understanding of how often the system completes a cycle every second.
The unit for angular frequency is radians per second (rad/s). Angular frequency is vitally linked to both the mass and the spring constant in simple harmonic systems through the formula:
  • \(\omega = \sqrt{\frac{k}{m}}\)
Here, \(k\) is the spring constant and \(m\) is the mass of the object in motion.
This equation indicates that the angular frequency increases with a stiffer spring (larger \(k\)) and decreases with an increase in mass (larger \(m\)). A higher angular frequency indicates more rapid oscillations, meaning the object returns to its starting position more quickly during each period of motion.
Oscillation Period
The oscillation period in simple harmonic motion is the time it takes for the system to complete one full cycle of motion and return to its initial position.
It is denoted by \(T\) and measured in seconds. The oscillation period is inversely related to the angular frequency, with the formula:
  • \(T = \frac{2\pi}{\omega}\)
This relationship indicates that as the angular frequency \(\omega\) increases, the period of oscillation \(T\) decreases, making the system oscillate more quickly.In practice, understanding the period assists in predicting the motion's timing, allowing precise analysis of the system's behavior over time.
For applications like clocks or measuring instruments, the period's consistency ensures accuracy in operation, given that each cycle takes the same time regardless of its amplitude in an ideal SHM system.

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Most popular questions from this chapter

A damped harmonic oscillator consists of a block \((m=2.00 \mathrm{~kg}),\) a spring \((k=10.0 \mathrm{~N} / \mathrm{m}),\) and a damping force \((F=-b v) .\) Initially, it oscillates with an amplitude of \(25.0 \mathrm{~cm}\) because of the damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. (a) What is the value of \(b ?\) (b) How much energy has been "lost" during these four oscillations?

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position \(y_{i}\) such that the spring is at its rest length. The object is then released from \(y_{i}\) and oscillates up and down, with its lowest position being \(10 \mathrm{~cm}\) below \(y_{i}\). (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is \(8.0 \mathrm{~cm}\) below the initial position? (c) An object of mass \(300 \mathrm{~g}\) is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) How far below \(y_{i}\) is the new equilibrium (rest) position with both objects attached to the spring?

A \(0.10 \mathrm{~kg}\) block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by $$ x=(10 \mathrm{~cm}) \cos [(10 \mathrm{rad} / \mathrm{s}) t+\pi / 2 \mathrm{rad}] $$ (a) What is the oscillation frequency? (b) What is the maximum speed acquired by the block? (c) At what value of \(x\) does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e) At what value of \(x\) does this occur? (f) What force, applied to the block by the spring, results in the given oscillation?

What is the maximum acceleration of a platform that oscillates at amplitude \(2.20 \mathrm{~cm}\) and frequency \(6.60 \mathrm{~Hz}\) ?

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