/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 A block of mass \(m\) is undergo... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 31

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 \(A / 2 ?\)

Short Answer

Expert verified
The displacement of the block from its equilibrium position when its speed is half its maximum speed is \(x = (A/\sqrt{2})\), which is larger than \(A/2\).

Step by step solution

01

Understand the relationship between displacement and velocity in SHM

In Simple Harmonic Motion (SHM), the velocity and displacement of the oscillating object are related by the equation \(v = \sqrt {k/m} \sqrt {A^2 - x^2}\), where \(A\) is the amplitude, \(x\) is the displacement from the equilibrium, \(k\) is the spring constant, \(m\) is the mass of the block, and \(v\) is the instantaneous speed of the block. The maximum speed \(v_{max}\) is when the block is at the equilibrium position \(x = 0\) and is given by \(v_{max} = \sqrt {k/m} \cdot A\)
02

Substitute the given values into the equation

Since we are asked to find the position when the speed \(v\) is half of its maximum speed \(v_{max}\), we can replace \(v\) with \(v_{max}/2\) in the equation. Hence the equation becomes \(v_{max}/2 = \sqrt {k/m} \sqrt {A^2 - x^2}\)
03

Solve the equation for displacement \(x\)

Square both sides of the equation to get rid of the square roots, you end up with \((v_{max}/2)^2 = (\sqrt {k/m})^2 (A^2 - x^2)\). Rewriting the right hand side equation, we have \(v_{max}^2/4 = k/m \cdot (A^2 - x^2)\). Substitute \(v_{max}\) with \(\sqrt {k/m} \cdot A\) into the equation to eliminate \(k\), you have \(\sqrt {k/m}^2 \cdot A^2/4 = A^2 - x^2\). Simplifying the equation yields \(A^2/4 = A^2 - x^2\) or \(x^2 = 3A^2/4\). Finally take square root gives displacement \(x = \sqrt {3/4} \cdot A\) or \(x = (A/\sqrt{2})\) which is larger than \(A/2\).
04

Compare to \(A / 2\)

\((A/\sqrt{2})\) is larger than \(A/2\). Hence, when the speed of the block is half of its maximum speed, its displacement from the equilibrium position is larger than half of its amplitude.

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Key Concepts

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

Displacement and Velocity in SHM
In the context of Simple Harmonic Motion (SHM), displacement and velocity are closely interconnected. The displacement (\(x\)) refers to how far the object is from its equilibrium position, while the velocity (\(v\)) is the speed of the object as it moves back and forth.
In SHM, the velocity at any point can be determined using the formula:\[ v = \sqrt{\frac{k}{m}} \cdot \sqrt{A^2 - x^2} \]where:
  • \(v\) is the instantaneous velocity.
  • \(k\) is the spring constant.
  • \(m\) is the mass of the object.
  • \(A\) is the amplitude of oscillation.
  • \(x\) is the displacement from equilibrium.
The velocity is highest at the equilibrium point (\(x = 0\)), and decreases as the object moves towards the extremities of its motion, where the displacement is greatest.
Maximum Speed in SHM
In SHM, the maximum speed is a crucial aspect to understand, as it occurs when the object passes through the equilibrium position.
This maximum speed (\(v_{max}\)) can be calculated using the relation:\[ v_{max} = \sqrt{\frac{k}{m}} \cdot A \]Here, the entire potential energy stored in the spring is converted into kinetic energy, translating to maximum velocity.
Beyond this point, as the object moves away, it begins to slow down due to the potential energy being stored back into the spring.
Amplitude of SHM
Amplitude (\(A\)) is the peak value of displacement in SHM. It reflects how far the object travels from its equilibrium position at its extreme points.
The amplitude is directly proportional to the energy stored in the motion, exemplifying the system's capability to perform work.
  • The amplitude is always a positive value.
  • It remains constant in ideal SHM, where no external forces (like friction) are present.
The amplitude plays a vital role in determining the maximum speed of the object, with \(v_{max} = \sqrt{\frac{k}{m}} \cdot A\), highlighting that a larger amplitude results in a higher maximum speed.
Equilibrium Position
The equilibrium position in SHM is the central point around which the object oscillates, denoted as \(x = 0\).
This position is significant as it is the point of maximum kinetic energy and zero potential energy. At the equilibrium, all forces balance out, causing neither net force nor acceleration.
When the object passes through this position, it reaches its maximum speed, as calculated by \(v_{max} = \sqrt{\frac{k}{m}} \cdot A\).
  • All subsequent motion is due to the restoring force that pulls or pushes the object back to equilibrium.
  • In reality, damping forces can cause the object to gradually come to a stop at this point if not continually energized.
Spring Constant
The spring constant, \(k\), is a measure of the stiffness of a spring. It represents how much force is needed to stretch or compress the spring by a unit length.
The value of \(k\) directly impacts the characteristics of SHM.
  • A larger \(k\) implies a stiffer spring, resulting in faster oscillations and higher frequency.
  • A smaller \(k\) indicates a more flexible spring, allowing for slower oscillations and lower frequencies.
\(k\) is instrumental in determining both the angular frequency of oscillation and the maximum speed, as it appears in the formula \(v_{max} = \sqrt{\frac{k}{m}} \cdot A\).

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Most popular questions from this chapter

A \(10.0 \mathrm{~kg}\) mass is traveling to the right with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) on a smooth horizontal surface when it collides with and sticks to a second \(10.0 \mathrm{~kg}\) mass that is initially at rest but is attached to one end of a light, horizontal spring with force constant \(170.0 \mathrm{~N} / \mathrm{m}\). The other end of the spring is fixed to a wall to the right of the second mass. (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

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

Object \(A\) has mass \(m_{A}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{A} .\) Object \(B\) has mass \(m_{B}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{B}\). The amplitude \(A_{A}\) for object \(A\) is twice the amplitude \(A_{B}\) for the motion of object \(B\). Also, \(m_{B}=4 m_{A}\) and \(k_{A}=9 k_{B}\). (a) What is the ratio of the maximum speeds of the two objects, \(v_{\max , A} / v_{\max , B} ?\) (b) What is the ratio of their maximum accelerations, \(a_{\max , A} / a_{\max , B} ?\)

A block of mass \(m\) is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring has force constant \(k\), and the amplitude of the \(\mathrm{SHM}\) is \(A\). The block has \(v=0,\) and \(x=+A\) at \(t=0 .\) It first reaches \(x=0\) when \(t=T / 4\) where \(T\) is the period of the motion. (a) In terms of \(T,\) what is the time \(t\) when the block first reaches \(x=A / 2 ?\) (b) The block has its maximum speed when \(t=T / 4\). What is the value of \(t\) when the speed of the block first reaches the value \(v_{\max } / 2 ?\) (c) Does \(v=v_{\max } / 2\) when \(x=A / 2 ?\)

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?
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