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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 \(\mathrm{SHM}\) 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
The maximum speed, speed at -0.015m, maximum acceleration, acceleration at -0.015m, and total mechanical energy of the glider can be calculated from the given information using the relevant equations of SHM and the properties of ideal springs. The results will be obtained after substituting the given values into the equations and performing the necessary arithmetic operations.

Step by step solution

01

Calculate the maximum speed

The maximum speed of the glider occurs when the glider is at the equilibrium position (x=0). It is given by \(v_{max} = A\sqrt{\frac{k}{m}}\), where A is the amplitude of the motion, k is the force constant and m is the mass of the glider. Substituting the given values, we get \(v_{max} = 0.040m \times \sqrt{\frac{450 N/m}{0.500 kg}}\)
02

Calculate the speed at x=-0.015m

The speed of the glider at a certain displacement x from the equilibrium position is given by the equation \(v = \sqrt{\frac{k}{m}(A^{2}-x^{2})}\), where k is the force constant, m is the mass of the glider, A is the amplitude and x is the displacement. Substituting the given values, we get \(v = \sqrt{\frac{450 N/m}{0.500 kg}\left((0.040 m)^{2} - (-0.015 m)^{2}\right)}\)
03

Determine the maximum acceleration of the glider

The maximum acceleration occurs at the positive and negative extremes of the glider motion, i.e. at the amplitudes +A and -A. This is given by the equation \(a_{max} = A\frac{k}{m}\), where A is the amplitude of the motion, k is the force constant and m is the mass of the glider. Substituting the given values, we get \(a_{max} = 0.040m \times \frac{450 N/m}{0.500 kg}\)
04

Calculate the acceleration at x=-0.015m

The acceleration of the glider at a certain displacement x from the equilibrium position is given by the equation \(a = -\frac{k}{m}x\), where k is the force constant, m is the mass of the glider and x is the displacement. Substituting the given values, we get \(a = -\frac{450 N/m}{0.500 kg}\times -0.015 m\)
05

Calculate the total mechanical energy of the glider

The total mechanical energy of an oscillating system in SHM is constant and is given by the equation \(E = \frac{1}{2}kA^{2}\), where k is the force constant and A is the amplitude. Substituting the given values, we get \(E = \frac{1}{2}\times 450 N/m \times (0.040 m)^{2}\)

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Key Concepts

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

Mechanical Energy
Mechanical energy in simple harmonic motion (SHM) is fascinating because it remains constant throughout the motion. As a system oscillates back and forth, energy transitions between kinetic and potential forms. This sums up to the total mechanical energy, which can be calculated by the equation: \[E = \frac{1}{2} k A^2\]Here, \(E\) represents total mechanical energy, \(k\) is the force constant, and \(A\) is the amplitude of the motion. The neat part about SHM is that energy conservation holds the entire system in order. At maximum displacement (amplitude), all the energy is potential, and at the equilibrium point, it is kinetic. This constant energy value allows us to easily predict and calculate various properties of the system.
Amplitude
Amplitude is a key concept in SHM because it defines the maximum extent of the oscillation from the equilibrium position. It's essentially how far the object moves in either direction from rest. Think of it as the 'reach' of the motion, which is an indicator of how much energy is in the system.In mathematical terms, the amplitude \(A\) is part of several formulas used in SHM calculations, including those for velocity and acceleration. For example, if you want to find out the maximum speed or maximum acceleration, the amplitude is a necessary component:
  • Maximum speed, \(v_{max} = A \sqrt{\frac{k}{m}}\)
  • Maximum acceleration, \(a_{max} = A \frac{k}{m}\)
These formulas show how amplitude directly affects the system's dynamics, giving a clearer picture of the oscillation's scale and energy.
Kinematics
Kinematics in the context of simple harmonic motion deals with the movement of the system through its cycle. It includes understanding how position, velocity, and acceleration relate to time and each other.Within SHM, the position of an object can be described as:\[x(t) = A \cos(\omega t + \phi)\]where \(x(t)\) is the position at time \(t\), \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant. Velocity and acceleration can be derived from this position function:
  • Velocity: \(v(t) = -A \omega \sin(\omega t + \phi)\)
  • Acceleration: \(a(t) = -A \omega^2 \cos(\omega t + \phi)\)
Mastering these relationships allows you to track the entire motion path of the object through its periodic cycles.
Force Constant
The force constant, often referred to as stiffness in springs, is denoted by \(k\) and is crucial in defining how a spring system behaves in simple harmonic motion. Essentially, it's a measure of the spring's resistance to being compressed or elongated. Mathematically, Hooke's Law explains this as:\[F = -kx\]where \(F\) is the force exerted by the spring, \(k\) is the force constant, and \(x\) is the displacement from the equilibrium position. In SHM, the force constant is significant because it influences both the speed and acceleration:
  • Maximum speed: \(v_{max} = A \sqrt{\frac{k}{m}}\)
  • Maximum acceleration: \(a_{max} = A \frac{k}{m}\)
Understanding the force constant helps determine how quickly and with what force the system can return to its equilibrium state.

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Most popular questions from this chapter

The jerk is defined to be the time rate of change of the acceleration. (a) If the velocity of an object undergoing SHM is given by \(v_{x}=-\omega A \sin (\omega t),\) what is the equation for the \(x\) -component of the jerk as a function of time? (b) What is the value of \(x\) for the object when the \(x\) -component of the jerk has its largest positive value? (c) What is \(x\) when the \(x\) -component of the jerk is most negative? (d) When it is zero? (e) If \(v_{x}\) equals \(-0.040 \mathrm{~s}^{2}\) times the \(x\) -component of the jerk for all \(t,\) what is the period of the motion?

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?

BIO Weighing a Virus. In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just \(30 \mathrm{nm}\) long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{V}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by \(f_{\mathrm{S}+\mathrm{V}} / f_{\mathrm{S}}=1 / \sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)},\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that it is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{~g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{~Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{~Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

DATA Experimenting with pendulums, you attach a light string to the ceiling and attach a small metal sphere to the lower end of the string. When you displace the sphere \(2.00 \mathrm{~m}\) to the left, it nearly touches a vertical wall; with the string taut, you release the sphere from rest. The sphere swings back and forth as a simple pendulum, and you measure its period \(T\). You repeat this act for strings of various lengths \(L\), each time starting the motion with the sphere displaced \(2.00 \mathrm{~m}\) to the left of the vertical position of the string. In each case the sphere's radius is very small compared with \(L\). Your results are given in the table: $$ \begin{array}{l|rrrrrrrr} \boldsymbol{L}(\mathbf{m}) & 12.00 & 10.00 & 8.00 & 6.00 & 5.00 & 4.00 & 3.00 & 2.50 & 2.30 \\ \hline \boldsymbol{T}(\mathbf{s}) & 6.96 & 6.36 & 5.70 & 4.95 & 4.54 & 4.08 & 3.60 & 3.35 & 3.27 \end{array} $$ (a) For the five largest values of \(L,\) graph \(T^{2}\) versus \(L\). Explain why the data points fall close to a straight line. Does the slope of this line have the value you expected? (b) Add the remaining data to your graph. Explain why the data start to deviate from the straight-line fit as \(L\) decreases. To see this effect more clearly, plot \(T / T_{0}\) versus \(L,\) where \(T_{0}=2 \pi \sqrt{L / g}\) and \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) (c) Use your graph of \(T / T_{0}\) versus \(L\) to estimate the angular amplitude of the pendulum (in degrees) for which the equation \(T=2 \pi \sqrt{L / g}\) is in error by \(5 \%\).

Consider the system of two blocks and a spring shown in Fig. \(\mathrm{P} 14.66 .\) The horizontal surface is friction less, but there is static friction between the two blocks. The spring has force constant \(k=150 \mathrm{~N} / \mathrm{m} .\) The masses of the two blocks are \(m=0.500 \mathrm{~kg}\) and \(M=4.00 \mathrm{~kg} .\) You set the blocks into motion by releasing block \(M\) with the spring stretched a distance \(d\) from equilibrium. You start with small values of \(d,\) and then repeat with successively larger values. For small values of \(d,\) the blocks move together in SHM. But for larger values of \(d\) the top block slips relative to the bottom block when the bottom block is released. (a) What is the period of the motion of the two blocks when \(d\) is small enough to have no slipping? (b) The largest value \(d\) can have and there be no slipping is \(d=8.8 \mathrm{~cm} .\) What is the coefficient of static friction \(\mu_{\mathrm{s}}\) between the surfaces of the two blocks?

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