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An object is undergoing SHM with period \(0.900 \mathrm{~s}\) and amplitude \(0.320 \mathrm{~m} .\) At \(t=0\) the object is at \(x=0.320 \mathrm{~m}\) and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x=0.320 \mathrm{~m}\) to \(x=0.160 \mathrm{~m}\) and (b) from \(x=0.160 \mathrm{~m}\) to \(x=0\)

Short Answer

Expert verified
The SHM object takes approximately 0.225 seconds to move from \(x=0.320m\) to \(x=0.160m\) and approximately 0.125 seconds to move from \(x=0.160m\) to \(x=0\).

Step by step solution

01

Define the Problem

An object undergoing SHM has a period of 0.900s and amplitude of 0.320m. The initial position at \(t=0\) is \(x = 0.320m\) and the object is momentarily at rest. The task is to find the time it takes to go from \(x=0.320m\) to \(x=0.160m\) and \(x=0.160m\) to \(x=0\).
02

Identify the Relevant Formulae

In SHM, the displacement \(x\) at time \(t\) can be represented by the equation \(x = A\cos(\omega t + \phi)\), where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. Based on the given conditions, we can calculate \(\omega\) with the formula \(\omega = \frac{2\pi}{T}\), where \(T\) is the period. Since the object is momentarily at rest at \(t=0\) and \(x=A\), we know that \(\phi = 0\). To find the time corresponding to a given displacement, we set \(x = A\cos(\omega t)\) and solve for \(t\).
03

Calculate the Angular Frequency \(\omega\)

Substitute the given period \(T = 0.900s\) into \(\omega = \frac{2\pi}{T}\) to find \(\omega\).
04

Calculate Time \(t\) to Move From \(x=0.320m\) to \(x=0.160m\)

Set \(x = A\cos(\omega t)\) to \(0.160m = 0.320m\cos(\omega t)\), divide through by \(0.320m\) to find the \(\cos(\omega t)\), use an arccos function to determine \(\omega t\), then divide \(\omega t\) by the calculated \(\omega\) to find \(t\).
05

Calculate Time \(t\) to Move From \(x=0.160m\) to \(x=0\)

Starting again at \(x = A\cos(\omega t)\), set \(x = 0\) and solve for \(t\) using the same technique as before. Remember the equation \(\cos(\omega t)\) has a periodicity, which results in multiple solutions for \(t\), but we need to find the smallest positive \(t\) which is during the first half-period.

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Key Concepts

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

Displacement in SHM
In Simple Harmonic Motion (SHM), displacement refers to how far an object is from its equilibrium or rest position. Displacement is typically denoted by the variable \(x\) and can be positive or negative, depending on the direction of movement from the equilibrium position. This is crucial in understanding how the object oscillates over time.

The formula used to calculate displacement in SHM is:
  • \( x(t) = A \cos(\omega t + \phi) \)
Here, \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. This equation highlights that the displacement of an object in SHM varies with time as a cosine function.

Understanding displacement in SHM helps us predict where the object is at any point in time during its motion.
Angular Frequency
Angular frequency, often represented by the Greek letter \(\omega\), plays a vital role in SHM. It measures how many oscillations occur in a unit of time and is related to the period \(T\) of the oscillation. The period is the time taken for one complete cycle of motion.

Angular frequency is calculated using the formula:
  • \( \omega = \frac{2\pi}{T} \)
Where \(T\) is the period of the oscillation. The factor \(2\pi\) arises because we consider one full circle of motion, similar to a circle having \(360^\circ\) or \(2\pi\) radians.

This concept ensures that we can accurately determine how quickly the object moves through its oscillatory path, affecting displacement and velocity.
Phase Constant
The phase constant, represented by \(\phi\) in the displacement equation of SHM, is crucial for determining the initial conditions of the motion. It indicates the starting point of the oscillation relative to a standard cosine function.

For example, when an object in SHM is at its maximum displacement at \(t=0\), as in the given exercise, the phase constant \(\phi\) is zero. This aligns the cosine function accurately with the motion's starting point.

Key points about the phase constant:
  • It's derived from initial conditions such as starting displacement and phase of motion.
  • Helps shift the cosine curve along the time axis to match the object's initial position.
By knowing \(\phi\), we ensure precise predictions of the object's position at future times.
Amplitude
Amplitude, denoted as \(A\), is a fundamental concept in SHM that indicates the maximum displacement of the object from its equilibrium position. In simple terms, it is the farthest distance the object travels in one direction before reversing its path.

Amplitude is always a positive value and remains constant throughout the motion, assuming no external forces alter the system.

Important characteristics of amplitude:
  • Reflects the energy level of the system—higher amplitudes represent more energy.
  • Determines the range of motion—objects in SHM oscillate back and forth between \(\pm A\).
Understanding the amplitude helps predict the maximum extent of an object's motion.
Cosine Function in SHM
In SHM, the cosine function is integral to describing the oscillatory motion. The displacement equation \(x(t) = A\cos(\omega t + \phi)\) clearly utilizes the cosine function to map position over time.

This function is periodic, meaning it repeats its values in regular intervals, which matches the nature of SHM perfectly. It cycles between -1 and 1, allowing us to calculate displacement at any point efficiently.

Applications of the cosine function in SHM include:
  • Modeling positional changes—capturing the back-and-forth motion typical of springs and pendulums.
  • Predicting time-based displacement—providing a mathematical map of the oscillation.
Through the cosine function, we can mathematically track the rhythmic movement of objects undergoing SHM, which is essential for predicting and understanding their motion.

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Most popular questions from this chapter

A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{~kg}\) and radius \(R=0.050 \mathrm{~m}\) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

A \(1.80 \mathrm{~kg}\) monkey wrench is pivoted \(0.250 \mathrm{~m}\) from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is \(0.940 \mathrm{~s}\). (a) What is the moment of inertia of the wrench about an axis through the pivot? (b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

A block of mass \(m\) is undergoing SHM on a horizontal, friction less surface while it is attached to a light, horizontal spring that has force constant \(k .\) The amplitude of the \(\mathrm{SHM}\) of the block is \(A .\) What is the distance \(|x|\) of the block from its equilibrium position when its speed \(v\) is half its maximum speed \(v_{\max } ?\) Is this distance larger or smaller than \(A / 2 ?\)

A block with mass \(m=0.300 \mathrm{~kg}\) is attached to one end of an ideal spring and moves on a horizontal friction less surface. The other end of the spring is attached to a wall. When the block is at \(x=+0.240 \mathrm{~m},\) its acceleration is \(a_{x}=-12.0 \mathrm{~m} / \mathrm{s}^{2}\) and its velocity is \(v_{x}=+4.00 \mathrm{~m} / \mathrm{s} .\) What are (a) the spring's force constant \(k ;\) (b) the amplitude of the motion; (c) the maximum speed of the block during its motion; and (d) the maximum magnitude of the block's acceleration during its motion?

A \(5.00 \mathrm{~kg}\) partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down \(0.100 \mathrm{~m}\) below its equilibrium position and released, it vibrates with a period of \(4.20 \mathrm{~s}\). (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is \(0.050 \mathrm{~m}\) above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point \(0.050 \mathrm{~m}\) below its equilibrium position to a point \(0.050 \mathrm{~m}\) above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

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