/*! 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 111 A \(2.0 \mathrm{~kg}\) block exe... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 111

A \(2.0 \mathrm{~kg}\) block executes SHM while attached to a horizontal spring of spring constant \(200 \mathrm{~N} / \mathrm{m}\). The maximum speed of the block as it slides on a horizontal frictionless surface is \(3.0 \mathrm{~m} / \mathrm{s}\). What are (a) the amplitude of the block's motion, (b) the magnitude of its maximum acceleration, and (c) the magnitude of its minimum acceleration? (d) How long does the block take to complete \(7.0 \mathrm{cy}\) cles of its motion?

Short Answer

Expert verified
(a) 0.3 m, (b) 30 m/s虏, (c) 0 m/s虏, (d) approx. 4.398 s.

Step by step solution

01

Understand the Concept of Maximum Speed in SHM

The maximum speed \(v_{max}\) in Simple Harmonic Motion (SHM) is given by the formula \(v_{max} = A\omega\), where \(A\) is the amplitude and \(\omega\) is the angular frequency. Our task is to find \(A\) first, so we'll rearrange the formula to \(A = \frac{v_{max}}{\omega}\).
02

Calculate the Angular Frequency

In SHM, the angular frequency \(\omega\) is related to the mass \(m\) and the spring constant \(k\) by the formula \(\omega = \sqrt{\frac{k}{m}}\). Here, \(k = 200 \mathrm{~N/m}\) and \(m = 2.0 \mathrm{~kg}\), so substitute these values:\[ \omega = \sqrt{\frac{200}{2.0}} = \sqrt{100} = 10 \, \mathrm{rad/s} \]
03

Determine the Amplitude

Now, substitute \(v_{max} = 3.0 \, \mathrm{m/s}\) and \(\omega=10 \, \mathrm{rad/s}\) into the formula for amplitude:\[ A = \frac{3.0}{10} = 0.3 \, \mathrm{m} \]
04

Calculate Maximum Acceleration

The maximum acceleration \(a_{max}\) in SHM is given by \(a_{max} = A\omega^2\). Substituting \(A = 0.3 \, \mathrm{m}\) and \(\omega = 10 \, \mathrm{rad/s}\) into the formula:\[ a_{max} = 0.3 \times 10^2 = 0.3 \times 100 = 30 \, \mathrm{m/s^2} \]
05

Determine Minimum Acceleration

In SHM, the minimum acceleration occurs at the mean position and is \(0 \, \mathrm{m/s^2}\).
06

Calculate the Time for Seven Cycles

The period \(T\) is given by \(T = \frac{2\pi}{\omega}\). For \(\omega = 10 \, \mathrm{rad/s}\),\[ T = \frac{2\pi}{10} = \frac{\pi}{5} \, \mathrm{s} \]To find the time for seven cycles, multiply the period by 7:\[ T_{7} = 7 \times \frac{\pi}{5} = \frac{7\pi}{5} \, \mathrm{s} \approx 4.398 \, \mathrm{s} \]

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Key Concepts

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

Amplitude
In Simple Harmonic Motion (SHM), the amplitude is a crucial concept as it represents the maximum extent of displacement of the oscillating object from its equilibrium position. Imagine the block attached to the spring. It moves back and forth between its maximum points on either side.
This maximum distance from the central position is the amplitude, often denoted as \( A \).

To calculate the amplitude in SHM, we use the relationship between maximum speed and angular frequency:
  • The formula is \( v_{max} = A\omega \).
  • Rearrange it to find amplitude: \( A = \frac{v_{max}}{\omega} \).
  • Substitute the known values, \( v_{max} = 3.0 \, \mathrm{m/s} \) and \( \omega = 10 \, \mathrm{rad/s} \).
  • The result is \( A = 0.3 \, \mathrm{m} \), meaning the block oscillates 0.3 meters from the center.
Understanding amplitude helps grasp how far the object travels in each cycle. It gives insight into the energy stored in the motion, indicating a higher energy if the amplitude is larger.
Angular Frequency
Angular frequency in SHM, denoted by \( \omega \), measures how quickly an object oscillates around its path.
It's analogous to the concept of frequency in circular motion, describing how fast the oscillation cycles are completed.

The angular frequency is calculated with:
  • Formula: \( \omega = \sqrt{\frac{k}{m}} \).
  • For our block, the spring constant \( k \) is 200 N/m and mass \( m \) is 2.0 kg.
  • Substituting these values yields \( \omega = \sqrt{\frac{200}{2}} = 10 \, \mathrm{rad/s} \).
The value of \( 10 \, \mathrm{rad/s} \) means the block completes its cycle quickly, showcasing a tight correlation between the stiffness of the spring and the oscillation speed.
This attribute helps to determine crucial aspects like amplitude and maximum acceleration in SHM.
Maximum Acceleration
Maximum acceleration in SHM is another essential concept as it indicates the highest rate of change in velocity an object experiences during its motion.
It is important because it tells us the force needed to achieve this, tying back to Newton鈥檚 second law.

To find this value:
  • Use the formula \( a_{max} = A\omega^2 \).
  • With amplitude \( A = 0.3 \, \mathrm{m} \) and angular frequency \( \omega = 10 \, \mathrm{rad/s} \),
  • The calculation is \( a_{max} = 0.3 \times 10^2 = 30 \, \mathrm{m/s^2} \).
This indicates that the acceleration reaches its peak at the maximum displacement from the equilibrium point.
Understanding maximum acceleration is crucial for imagining how forces act on objects in SHM and ensures things like safety and durability in mechanical systems.

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Most popular questions from this chapter

A \(0.12 \mathrm{~kg}\) body undergoes simple harmonic motion of amplitude \(8.5 \mathrm{~cm}\) and period \(0.20 \mathrm{~s}\). (a) What is the magnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the spring constant?

The balance wheel of an old-fashioned watch oscillates with angular amplitude \(\pi\) rad and period \(0.500 \mathrm{~s}\). Find (a) the maximum angular speed of the wheel, (b) the angular speed at displacement \(\pi / 2 \mathrm{rad}\), and \((\mathrm{c})\) the magnitude of the angular acceleration at displacement \(\pi / 4\) rad.

A vertical spring stretches \(9.6 \mathrm{~cm}\) when a \(1.3 \mathrm{~kg}\) block is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional \(5.0 \mathrm{~cm}\) downward and released from rest. Find the (b) period, (c) frequency, (d) amplitude, and (e) maximum speed of the resulting SHM.

A block is in SHM on the end of a spring, with position given by \(x=x_{m} \cos (\omega t+\phi)\). If \(\phi=\pi / 5 \mathrm{rad}\), then at \(t=0\) what percentage of the total mechanical energy is potential energy?

The suspension system of a \(2000 \mathrm{~kg}\) automobile "sags" \(10 \mathrm{~cm}\) when the chassis is placed on it. Also, the oscillation amplitude decreases by \(50 \%\) each cycle. Estimate the values of (a) the spring constant \(k\) and (b) the damping constant \(b\) for the spring and shock absorber system of one wheel, assuming each wheel supports \(500 \mathrm{~kg}\).
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