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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

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 \(5 \%\).

In a physics lab, you attach a \(0.200 \mathrm{~kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is \(2.60 \mathrm{~s}\). Find the spring's force constant.

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 \(g=3.71 \mathrm{~m} / \mathrm{s}^{2} ?\)

A thrill-seeking cat with mass \(4.00 \mathrm{~kg}\) is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is \(0.050 \mathrm{~m},\) and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; (b) at its lowest point; (c) at its equilibrium position.

An object is undergoing SHM with period \(0.900 \mathrm{~s}\) and amplitude \(0.320 \mathrm{~m} .\) At \(t=0\) the object is at \(x=0.320 \mathrm{~m}\) and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x=0.320 \mathrm{~m}\) to \(x=0.160 \mathrm{~m}\) and (b) from \(x=0.160 \mathrm{~m}\) to \(x=0\)
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