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A 0.500 \(\mathrm{kg}\) glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) , undergoes SHM with an amplitude of 0.040 \(\mathrm{m}\) . Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m} ;\) (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x=-0.015 \mathrm{m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

Short Answer

Expert verified
(a) 1.2 m/s; (b) 1.113 m/s; (c) 36 m/s²; (d) 13.5 m/s²; (e) 0.36 J.

Step by step solution

01

Identify Given Values

First, we need to list out the given values in the problem. - Mass of glider, \( m = 0.500 \; \text{kg} \)- Force constant of spring, \( k = 450 \; \text{N/m} \)- Amplitude of SHM, \( A = 0.040 \; \text{m} \)- Position \( x = -0.015 \; \text{m} \) for specific calculations.
02

Calculate Maximum Speed

In SHM, the maximum speed \( v_{max} \) is given by the formula \( v_{max} = A \omega \), where \( \omega = \sqrt{\frac{k}{m}} \) is the angular frequency.First, calculate \( \omega \):\[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{450}{0.500}} = 30 \; \text{rad/s} \]Now, calculate \( v_{max} \):\[ v_{max} = A \omega = 0.040 \times 30 = 1.2 \; \text{m/s} \]
03

Calculate Speed at x = -0.015 m

The speed \( v \) at any displacement \( x \) in SHM is determined by the equation:\[ v = \omega \sqrt{A^2 - x^2} \]Substitute known values:\[ v = 30 \sqrt{0.040^2 - (-0.015)^2} \]\[ v = 30 \sqrt{0.0016 - 0.000225} \]\[ v = 30 \sqrt{0.001375} \]\[ v \approx 30 \times 0.0371 = 1.113 \; \text{m/s} \]
04

Calculate Maximum Acceleration

The magnitude of the maximum acceleration \( a_{max} \) is given by \( a_{max} = A \omega^2 \).Calculate \( a_{max} \):\[ a_{max} = 0.040 \times 30^2 = 0.040 \times 900 = 36 \; \text{m/s}^2 \]
05

Calculate Acceleration at x = -0.015 m

The acceleration \( a \) at any displacement \( x \) is given by the equation:\[ a = -\frac{k}{m}x \]Calculate \( a \):\[ a = -\frac{450}{0.500} \times (-0.015) = 900 \times 0.015 \]\[ a = 13.5 \; \text{m/s}^2 \]
06

Calculate Total Mechanical Energy

The total mechanical energy \( E \) in SHM is constant and given by:\[ E = \frac{1}{2} k A^2 \]Calculate \( E \):\[ E = \frac{1}{2} \times 450 \times 0.040^2 \]\[ E = 0.5 \times 450 \times 0.0016 \]\[ E = 0.36 \; \text{J} \]

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Key Concepts

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

Maximum Speed
In Simple Harmonic Motion (SHM), an object like our glider reaches its maximum speed when it is passing through the equilibrium position. This is because, at this point, the potential energy is completely converted into kinetic energy. The formula to determine this maximum speed, often denoted as \( v_{max} \), is given by:
  • \( v_{max} = A \omega \)
where \( A \) is the amplitude of motion, and \( \omega \) represents the angular frequency.

The angular frequency \( \omega \) can be calculated using the formula:
  • \( \omega = \sqrt{\frac{k}{m}} \)
where \( k \) is the spring constant, and \( m \) is the mass of the glider.

From the given values, \( A = 0.040 \, \text{m} \) and \( \omega = 30 \, \text{rad/s} \), we find that the maximum speed \( v_{max} = 1.2 \, \text{m/s} \). This maximum speed is the highest speed the glider reaches as it oscillates back and forth.
Angular Frequency
Angular frequency is a key concept in SHM, determining how fast the oscillations occur. It is denoted by \( \omega \) and is measured in radians per second (rad/s). Angular frequency is related to the spring constant \( k \) and the mass \( m \) by the formula:
  • \( \omega = \sqrt{\frac{k}{m}} \)
This equation shows the intrinsic connection between the stiffness of the spring and the inertia of the mass. For our glider, with \( k = 450 \, \text{N/m} \) and \( m = 0.500 \, \text{kg} \), the angular frequency \( \omega \) is calculated as \( 30 \, \text{rad/s} \).

This value is important because it influences both the maximum speed and the maximum acceleration of the glider during its oscillation.
Mechanical Energy
In SHM, the mechanical energy remains constant throughout the motion, consisting of both kinetic and potential energy. The total mechanical energy \( E \) can be expressed as:
  • \( E = \frac{1}{2} k A^2 \)
where \( k \) is the spring constant and \( A \) is the amplitude.

This energy expression highlights that the energy is proportional to both the square of the amplitude and the stiffness of the spring. For the glider with \( A = 0.040 \, \text{m} \) and \( k = 450 \, \text{N/m} \), we find the total mechanical energy is \( 0.36 \, \text{J} \). This energy value is maintained throughout the oscillation, ensuring that the system is conservative with interchangeable kinetic and potential energy.
Acceleration
Acceleration in Simple Harmonic Motion is tied to the position of the oscillating object and the spring constant. The maximum acceleration is experienced at the extreme positions of motion, calculated using:
  • \( a_{max} = A \omega^2 \)
For the given values, \( a_{max} = 36 \, \text{m/s}^2 \), demonstrating how the acceleration peaks when the glider is furthest from the equilibrium point.

Moreover, the acceleration at any position \( x \) can be computed as:
  • \( a = -\frac{k}{m}x \)
At \( x = -0.015 \, \text{m} \), the acceleration becomes \( 13.5 \, \text{m/s}^2 \). These calculations show that acceleration is greatest at the turning points and changes direction as the glider moves through the equilibrium point, following Hooke's Law principles.

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Most popular questions from this chapter

A machine part is undergoing SHM with a frequency of 5.00 Hz and amplitude 1.80 \(\mathrm{cm} .\) How long does it take the part to go from \(x=0\) to \(x=-1.80 \mathrm{cm} ?\)

Jerk. A guitar string vibrates at a frequency of 440 \(\mathrm{Hz}\) . A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

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 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 uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 \(\mathrm{kg}\) , and a \(175-\mathrm{kg}\) sack of gravel sits on the middle of it. The beam is oscillating in SHM, with an amplitude of 40.0 \(\mathrm{cm}\) and a frequency of 0.600 cycles \(/ \mathrm{s}\). (a) The sack of gravel falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the gravel instead falls off when the beam has its maximum speed, what are the frequency and amplitude of the subsequent SHM of the beam?

An approximation for the potential energy of a KCl molecule is \(U=A\left[\left(R_{0}^{7} / 8 r^{8}\right)-1 / r\right],\) where \(R_{0}=2.67 \times 10^{-10} \mathrm{m}\) and \(A=2.31 \times 10^{-28} \mathrm{J} \cdot \mathrm{m}\) . Using this approximation: (a) Show that the radial component of the force on each atom is \(F_{r}=A\left[\left(R_{0}^{7} / r^{9}\right)-1 | r^{2}\right] .(b)\) Show that \(R_{0}\) is the equilibrium separation. (c) Find the minimum potential energy. (d) Use \(r=R_{0}+x\) and the first two terms of the binomial theorem (Eq. 13.28 ) to show that \(F_{r} \approx-\left(7 A / R_{0}^{3}\right) x,\) so that the molecule's force constant is \(k=7 A / R_{0}^{3} .\) (e) With both the \(K\) and \(C l\) atoms vibrating in opposite directions on opposite sides of the molecule's center of mass, \(m_{1} m_{2} /\left(m_{1}+m_{2}\right)=3.06 \times 10^{-26} \mathrm{kg}\) is the mass to use in calculating the frequency (see Problem 13.86\() .\) Calculate the frequency of small-amplitude vibrations.

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