/*! 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 15 A block of mass \(m\) is undergo... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 15

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 \(x=A / 2 ?\)

Short Answer

Expert verified
(a) The time \(t\) when the block first reaches \(x = A/2\) is \(t = (T/2蟺)arccos(1/2)\). (b) The time \(t\) when the speed of the block first reaches the value \(v_{max}/2\) is \(t = (T/2蟺)arcsin(1/2)\). (c) The block may not necessarily reach \(v_{max}/2\) at \(x = A/2\), they can happen at different times

Step by step solution

01

Equation of motion for SHM

The position of the block as a function of time is given by the relation \(x(t) = Acos(蠅t+蠁)\) where \(蠅 = 2蟺/T\) is the angular frequency, \(蠁\) is the phase constant, \(A\) is the amplitude and \(T\) is the period.
02

Determine initial conditions

Given that the block is initially at \(x=A\) (at \(t=0\)), the phase constant \(蠁\) can be found. We plug these values into the equation of motion, getting \(A = A*cos(蠁) \), hence, \(蠁 = 0\). The equation of motion simplifies to \(x(t) = Acos(蠅t)\).
03

Calculate time for block to reach \(x = A/2\) for the first time

From equation (2), we can write for \(x = A/2\): \(A/2 = Acos(蠅t)\). Solving for \(t\), we get \(t = (1/蠅)arccos(1/2) = (T/2蟺)arccos(1/2)\)
04

Formulate the equation for velocity of block

The velocity of the block can be calculated from the derivative of the position equation. \(v(t) = -蠅Asin(蠅t) \)
05

Calculate time for velocity to reach \(v_{max} / 2\) for the first time

Given that \(v_{max} = | -蠅A | = 蠅A\), hence, for \(v(t) = v_{max}/2\), we have \(蠅A/2 = -蠅Asin(蠅t)\). Solving for \(t\), we get \(t = (1/蠅)arcsin(1/2) = (T/2蟺)arcsin(1/2)\)
06

Determine if velocity is \(v_{max}/2\) when \(x = A/2\)

It is not necessarily true that the block has velocity \(v_{max}/2\) when \(x = A/2\). These events may happen at different instances.

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Key Concepts

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

Equation of Motion
In simple harmonic motion (SHM), the equation of motion beautifully describes how an object moves back and forth along the same path. This equation is given by \[x(t) = A\cos(蠅t + 蠁)\]where:
  • \(x(t)\): Position of the object as a function of time
  • \(A\): Amplitude, the maximum extent of oscillation from the equilibrium position.
  • \(蠅\): Angular frequency, related to the speed of oscillation.
  • \(蠁\): Phase constant, which determines where in its cycle the motion begins.
This formula can be used to predict the position of an oscillating object at any given moment in time.
Angular Frequency
Angular frequency is an essential component in the study of SHM, representing how quickly something oscillates. The formula for angular frequency is:\[ 蠅 = \frac{2蟺}{T} \]where:
  • \(蠅\): Angular frequency, often measured in radians per second.
  • \(T\): Period, or the time taken for one complete cycle of motion.
Angular frequency helps in understanding not just the speed but also the energy levels in the system. The larger the angular frequency, the faster the object oscillates between its extremes.
Phase Constant
The phase constant \(蠁\) is instrumental in specifying the initial conditions of oscillation. Much like setting a starting point on a clock face, it helps determine where the oscillation begins. For our situation, the object starts at its extreme position \(x = A\) at time \(t = 0\). By plugging these values into the SHM equation, we find that the phase constant is \(蠁 = 0\).This indicates that, at time zero, the oscillation is exactly at the point where the cosine curve itself would start.
Velocity in SHM
Velocity in simple harmonic motion is a measure of how fast the object moves through its oscillation. It's found by differentiating the position equation:\[v(t) = -蠅A\sin(蠅t)\]where:
  • \(v(t)\): Velocity at time \(t\).
  • \(-蠅A\): The maximum possible velocity, occurring when \(\sin(蠅t)\) equals one.
The negative sign indicates that the direction of velocity is opposite to that of displacement at maximum speed. As in the example exercise, if we need the time at which the velocity first reaches a certain fraction of its maximum value, we can solve \[|v(t)| = \frac{蠅A}{2}\] for \(t\). Remember, velocity can reach a stated fraction of its maximum at different positions on the displacement path, such as \(x = A/2\) or others.

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Most popular questions from this chapter

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 \(x=-1.50 \mathrm{~cm}\)

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 Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where \(g=3.71 \mathrm{~m} / \mathrm{s}^{2} ?\)

\(\mathrm{A}\) 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?

A \(1.50 \mathrm{~kg},\) horizontal, uniform tray is attached to a vertical ideal spring of force constant \(185 \mathrm{~N} / \mathrm{m}\) and a \(275 \mathrm{~g}\) metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A,\) which is \(15.0 \mathrm{~cm}\) below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?
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