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Chapter 14: Problem 16

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is \(0.090 \mathrm{~m},\) it takes the block \(2.70 \mathrm{~s}\) to travel from \(x=0.090 \mathrm{~m}\) to \(x=-0.090 \mathrm{~m} .\) If the amplitude is doubled, to \(0.180 \mathrm{~m},\) how long does it take the block to travel (a) from \(x=0.180 \mathrm{~m}\) to \(x=-0.180 \mathrm{~m}\) and (b) from \(x=0.090 \mathrm{~m}\) to \(x=-0.090 \mathrm{~m} ?\)

Short Answer

Expert verified
The time taken for the block to travel (a) from \(x=0.180m\) to \(x=-0.180m\) and (b) from \(x=0.090m\) to \(x=-0.090m\) after the amplitude is doubled, is \(2.70s\) in both cases.

Step by step solution

01

Find the initial half period

The time to travel from \(x=0.090m\) to \(x=-0.090m\) is 2.70s, which is the initial half period. So, the initial half period, \(T1/2 = 2.70s\).
02

Determine the time for (a)

When the amplitude is doubled to \(0.180m\), the period of the motion does not change because the period of motion relies on mass \(m\) and spring constant \(k\), but not amplitude. The time to travel from \(x=0.180m\) to \(x=-0.180m\) will be equal to the initial half period, \(T1/2 = 2.70s\).
03

Determine the time for (b)

The time to travel from \(x=0.090m\) to \(x=-0.090m\), even after the amplitude is raised to \(0.180m\), doesn't change because the period doesn't change. So, it will also be equal to the initial half period, \(T1/2 = 2.70s\).

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Key Concepts

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

Period of Motion
In simple harmonic motion (SHM), the period of motion is a key concept. The period is the time it takes for a complete cycle of motion to occur. This means it goes from one point, through a complete back-and-forth swing, and returns to that point. In equation form, the period \( T \) of a mass-spring system is given by:\[T = 2\pi \sqrt{\frac{m}{k}}.\]Here, \( m \) is the mass attached to the spring, and \( k \) is the spring constant. The remarkable aspect of the period is that it does not depend on the amplitude of the motion. This is why when the amplitude changes, as in our exercise scenario, the period remains unchanged.
  • The period purely depends on the mass and the spring constant.
  • With higher mass, the period increases making the motion slower.
  • A larger spring constant indicates a stiffer spring, reducing the period and accordingly speeds up the motion.
Understanding the period is crucial because it helps predict the timing of each oscillation cycle in simple harmonic motion.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It is a fundamental part of Hooke's Law, which describes the behavior of springs. According to Hooke's Law:\[F = -kx,\]where \( F \) is the force applied to the spring and \( x \) is the displacement from the spring's equilibrium position. The spring constant \( k \) quantifies how much force is needed to stretch or compress the spring by a unit length.
  • A high spring constant means the spring is very stiff and resists deformation.
  • A low spring constant indicates a more flexible spring.
In our context of simple harmonic motion, the spring constant plays a vital role in determining the period of motion, as described in the period formula. It doesn't directly affect amplitude effects, but it fundamentally affects how quickly or slowly the system oscillates.
Amplitude
Amplitude in simple harmonic motion refers to the maximum extent of vibration or displacement from an equilibrium position. It's a measure of the energy within the system. In the context of a spring-mass system, the amplitude is the farthest distance the mass moves from its rest position.
  • A larger amplitude indicates more energy is stored in the system.
  • The system travels farther and its kinetic and potential energy peaks are higher.
  • However, amplitude does not affect the period of motion. This is a unique characteristic that differentiates simple harmonic motion from other types of movement.
In the example exercise, altering the amplitude from \(0.090 \mathrm{~m}\) to \(0.180 \mathrm{~m}\) does not affect the time taken for each half period. This highlights that in simple harmonic motion, while amplitude changes the path length, the timing is consistent.

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Most popular questions from this chapter

Quantum mechanics is used to describe the vibrational motion of molecules, but analysis using classical physics gives some useful insight. In a classical model the vibrational motion can be treated as SHM of the atoms connected by a spring. The two atoms in a diatomic molecule vibrate about their center of mass, but in the molecule HI, where one atom is much more massive than the other, we can treat the hydrogen atom as oscillating in SHM while the iodine atom remains at rest. (a) A classical estimate of the vibrational frequency is \(f=7 \times 10^{13} \mathrm{~Hz}\). The mass of a hydrogen atom differs little from the mass of a proton. If the HI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring? (b) The vibrational energy of the molecule is measured to be about \(5 \times 10^{-20} \mathrm{~J}\). In the classical model, what is the maximum speed of the H atom during its SHM? (c) What is the amplitude of the vibrational motion? How does your result compare to the equilibrium distance between the two atoms in the HI molecule, which is about \(1.6 \times 10^{-10} \mathrm{~m} ?\)

\(\mathrm{A}\) mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\) ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length \(50.0 \mathrm{~cm} .\) She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

DATA You hang various masses \(m\) from the end of a vertical, \(0.250 \mathrm{~kg}\) spring that obeys Hooke's law and is tapered, which means the diameter changes along the length of the spring. since the mass of the spring is not negligible, you must replace \(m\) in the equation \(T=2 \pi \sqrt{m / k}\) with \(m+m_{\text {eff }},\) where \(m_{\text {eff }}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass \(m\) and measure the time for 10 complete oscillations, obtaining these data: $$ \begin{array}{l|lcccc} \boldsymbol{m}(\mathbf{k g}) & 0.100 & 0.200 & 0.300 & 0.400 & 0.500 \\ \hline \text { Time (s) } & 8.7 & 10.5 & 12.2 & 13.9 & 15.1 \end{array} $$ (a) Graph the square of the period \(T\) versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring's effective mass. (d) What fraction is \(m_{\text {eff }}\) of the spring's mass? (e) If a \(0.450 \mathrm{~kg}\) mass oscillates on the end of the spring, find its period, frequency, and angular frequency.

A machine part is undergoing SHM with a frequency of \(4.00 \mathrm{~Hz}\) and amplitude \(1.80 \mathrm{~cm} .\) How long does it take the part to go from \(x=0\) to \(x=-1.80 \mathrm{~cm} ?\)
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