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An object is undergoing SHM with period \(0.300 \mathrm{~s}\) and amplitude \(6.00 \mathrm{~cm} .\) At \(t=0\) the object is instantaneously at rest at \(x=6.00 \mathrm{~cm}\). Calculate the time it takes the object to go from \(x=6.00 \mathrm{~cm}\) to \(x=-1.50 \mathrm{~cm}\)

Short Answer

Expert verified
The time it would take for the object to go from \(6.00 \mathrm{cm}\) to \(-1.50 \mathrm{cm}\) is approximately 0.183 seconds.

Step by step solution

01

Find the phase constant

The equation of motion for SHM is \(x=A\cos(\omega t+\phi)\). Since at \(t=0, x=6.00 \mathrm{cm}\), we can find \(\phi\). Therefore, rearranging for \(\phi\), we find that \(\phi=\arccos\left(\frac{x(t=0)}{A}\right)=\arccos\left(\frac{6 \mathrm{cm}}{6 \mathrm{cm}}\right)=\arccos(1)=0\). So the phase constant \(\phi = 0\).
02

Adjust the SHM formula for our specific case

From step 1, we have the equation of motion as \(x=6\cos\left(\frac{2\pi}{0.3}t\right)\). We now know that the object goes from \(x=6.00 \mathrm{cm}\) to \(x=-1.50 \mathrm{cm}\). So, we substitute \(x=-1.50 \mathrm{cm}\) into the equation giving: \(-1.50 = 6\cos\left(\frac{2\pi}{0.3}t\right)\).
03

Solve for time (t)

Rearranging step 2 for \(t\) gives: \(t = \frac{0.3}{2\pi}\times\arccos\left(\frac{-1.50 \mathrm{cm}}{6.00 \mathrm{cm}}\right)\) which equals approximately 0.183 seconds.

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Key Concepts

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

SHM Time Calculation
Understanding how to calculate the time taken for an object to travel from one point to another during simple harmonic motion (SHM) is crucial for analyzing oscillatory systems. The time calculation in SHM involves finding the moment when the object reaches a specific position as it moves back and forth around the equilibrium position.

From the original exercise, the object's time to travel from 6.00 cm to -1.50 cm is found using the rearranged SHM formula and solving for the time variable. A key aspect here is to ensure the units for time, position, and angular frequency are consistent to avoid any computational errors. It's important to note that the time calculated is only a fraction of the total period of the SHM, demonstrating a single snapshot within the full oscillatory cycle.

The time calculation often involves inverse trigonometric functions like arccos, arcsin, or arctan, since SHM can be described using trigonometric functions. In the exercise, for instance, arccos is used to find the time at which the object is at -1.50 cm. This involves the inverse cosine of the ratio of the desired position to the amplitude, followed by adjusting for the period and angular frequency of the motion.
Phase Constant
The phase constant, symbolized by \(\phi\), is a term that appears in the equations of motion for SHM and represents the initial angle (or phase) of the oscillating particle at time zero. It dictates where in its cycle the particle begins and can be critical for understanding the system's behavior at any given time.

In the context of the presented problem, determining the phase constant involved utilizing the initial conditions given. The object was at its maximum amplitude and at rest at time zero, which only occurs at two points in a cosine function: when \(\phi = 0\) or \(\phi = \pi\). Since the object was at a positive amplitude, \(\phi = 0\) was the correct solution. This phase constant helps define the starting point of the motion in the equation.

The equation \(x=A\cos(\omega t+\phi)\) allows for the calculation of position at any time. When the phase constant is zero, it indicates that the motion starts from the maximum amplitude. If the phase constant was \(\pi\), it would indicate a start from the negative amplitude. The phase constant is thus an essential element in defining the unique characteristics of an oscillatory system's motion.
Equation of Motion for SHM
The equation of motion for simple harmonic motion (SHM) provides a mathematical framework for predicting the future positions of an object oscillating at a harmonic frequency. It is typically represented as \(x=A\cos(\omega t+\phi)\) or \(x=A\sin(\omega t+\phi)\), where:\
    \
  • \(x\) is the displacement from the equilibrium position\
  • \
  • \(A\) is the amplitude, the maximum displacement\
  • \
  • \(\omega\) is the angular frequency, related to the period \(T\) by \(\omega = \frac{2\pi}{T}\)\
  • \
  • \(\phi\) is the phase constant\
  • \
  • \(t\) is the time\
  • \
\
These equations are fundamental in SHM as they describe the cyclical nature of the motion. The choice between cosine and sine generally depends on the initial conditions of the system.

In the exercise solution, the angular frequency is calculated using the given period and the amplitude is provided. The cosine function represents the SHM equation because the initial displacement was at the amplitude, which corresponds to a cosine curve starting at its peak value. This equation of motion is the key tool to understand how the object will move over time, with phase constant as the starting reference point.

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Most popular questions from this chapter

\(\mathrm{A}\) block with mass \(m\) is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring that has force constant \(k\). You use motion sensor equipment to measure the maximum speed of the block during its oscillations. You repeat the measurement for the same spring and blocks of different masses while keeping the amplitude \(A\) at a constant value of \(12.0 \mathrm{~cm}\). You plot your data as \(v_{\max }^{2}\) versus \(1 / m\) and find that the data lie close to a straight line that has slope \(8.62 \mathrm{~N} \cdot \mathrm{m} .\) What is the force constant \(k\) of the spring?

\(\mathrm{A}\) mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\) ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

SHM of a Butcher's Scale. A spring of negligible mass and force constant \(k=400 \mathrm{~N} / \mathrm{m}\) is hung vertically, and a \(0.200 \mathrm{~kg}\) pan is suspended from its lower end. A butcher drops a \(2.2 \mathrm{~kg}\) steak onto the pan from a height of \(0.40 \mathrm{~m} .\) The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

A block with mass \(M\) rests on a friction less surface and is connected to a horizontal spring of force constant \(k .\) The other end of the spring is attached to a wall (Fig. \(\mathbf{P 1 4 . 6 6 )} .\) A second block with mass \(m\) rests on top of the first block. The coefficient of static friction between the blocks is \(\mu_{\mathrm{s}}\). Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of \(0.450 \mathrm{~N} \cdot \mathrm{m} / \mathrm{rad}\). You twist the part a small amount about this axis and let it go, timing 165 oscillations in \(265 \mathrm{~s}\). What is its moment of inertia?

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