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A mass \(M=0.460 \mathrm{~kg}\) moves with an initial speed \(v=3.20 \mathrm{~m} / \mathrm{s}\) on a level frictionless air track. The mass is initially a distance \(D=0.250 \mathrm{~m}\) away from a spring with \(k=\) \(840 \mathrm{~N} / \mathrm{m}\), which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance \(d\), before reversing direction. After bouncing off the spring the mass travels with the same speed \(v\), but in the opposite dircction. a) Determine the maximum distance that the spring is compressed. b) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)

Short Answer

Expert verified
Based on the given problem: a) The maximum distance that the spring is compressed (d) is approximately 0.0930 meters. b) The total elapsed time until the mass returns to its starting point is approximately 0.230 seconds.

Step by step solution

01

Apply the conservation of energy principle

The initial kinetic energy of the mass is equal to the potential energy stored in the spring at maximum compression. This can be written as: $$ \frac{1}{2}Mv^2 = \frac{1}{2}kd^2 $$
02

Solve for d

We can solve the equation above for d: $$ d=\sqrt{\frac{Mv^2}{k}} $$
03

Calculate d

Now we can plug in the values given in the problem: $$ d = \sqrt{\frac{(0.460\,\text{kg})(3.20\,\text{m/s})^2}{840\,\text{N/m}}} $$ Calculate the numerical value of d: $$ d \approx 0.0930\,\text{m} $$ #b) Find the total elapsed time until the mass returns to its starting point.#
04

Calculate the time spent compressing the spring

When the mass is compressing the spring, it undergoes simple harmonic motion. The period of simple harmonic motion can be calculated using the formula: $$ T = 2\pi\sqrt{\frac{M}{k}} $$ The mass undergoes 1/2 of a full cycle while compressing the spring, so the time spent compressing is 1/2 the period: $$ t_\text{compress}=\frac{1}{2} \cdot 2\pi\sqrt{\frac{M}{k}} = \pi\sqrt{\frac{M}{k}} $$
05

Calculate the time spent moving back to the starting point

Once the mass bounces back, it travels back to the starting point with the same speed as it initially had. We can calculate the time it takes for the mass to travel back using the formula: $$ t_\text{return} = \frac{D}{v} $$
06

Calculate the total elapsed time

Now we can calculate the total elapsed time by adding the time spent compressing the spring and the time spent moving back to the starting point: $$ t_\text{total} = t_\text{compress}+t_\text{return} $$ Plugging in the values we found: $$ t_\text{total} = \pi\sqrt{\frac{(0.460\,\text{kg})}{(840\,\text{N/m})}} + \frac{(0.250\,\text{m})}{(3.20\,\text{m/s})} $$ Calculate the numerical value of the total elapsed time: $$ t_\text{total} \approx 0.230\,\text{s} $$

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics, which states that energy cannot be created or destroyed, only transformed from one form to another. This idea is crucial in understanding the motion of the mass and spring system. Initially, all the energy in the system is kinetic, due to the movement of the mass along the air track. When the mass hits the spring, this kinetic energy is gradually converted into potential energy as the spring compresses.

At the point of maximum compression, all initial kinetic energy has been transformed into potential energy stored in the spring. This is described mathematically by the equation:
  • Initial Kinetic Energy = Potential Energy at Maximum Compression
  • \[\frac{1}{2}Mv^2 = \frac{1}{2}kd^2\]
This equation serves to find the maximum compression of the spring, showing how the energy transformations help us understand the physical behaviors of objects.
Kinetic Energy
Kinetic energy represents the energy an object possesses due to its motion. In our scenario, the mass has kinetic energy as it moves along the track with a certain velocity. The formula to calculate kinetic energy is:
  • \[KE = \frac{1}{2}mv^2\]
  • Where \(m\) is the mass, and \(v\) is its velocity.
At the beginning of the problem, the mass has a kinetic energy based on its motion before it makes contact with the spring. This kinetic energy is completely converted into potential energy at the maximum compression of the spring. It is beneficial for students to recognize how kinetic energy varies with different velocities and masses, and its role in dynamics and collisions.
Potential Energy
Potential energy in the context of springs is the energy stored within the spring when it is compressed or stretched. In this problem, potential energy is maximized when the spring reaches its greatest compression. We use Hooke's law equation to find the energy stored:
  • Potential Energy = \( \frac{1}{2}kd^2 \)
  • \(k\) is the spring constant, \(d\) is the distance the spring is compressed.
We see that potential energy depends significantly on how much the spring is displaced. In the end, this stored energy transforms back to kinetic energy as the mass reverses its direction, allowing it to move back along the track. Comprehending these energy exchanges sharpens our analysis of mechanical systems.
Spring Constant
The spring constant, denoted by \(k\), is a measure of a spring's stiffness. It describes the relationship between the force applied to a spring and the displacement produced. Its units are newtons per meter (N/m).
  • The equation \(F = kx\) illustrates this relationship, where \(F\) is the force applied, and \(x\) is the displacement.
  • A larger \(k\) value signifies a stiffer spring, requiring more force to achieve the same displacement when compared to a spring with a smaller \(k\).
In our problem, the spring constant is a crucial value because it determines the spring's resistance to the force of the mass. Understanding the spring constant gives insight into how much the spring will compress under certain forces. This fundamental concept is integral for designing systems where springs can absorb energy or apply controlled forces.

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Most popular questions from this chapter

If you kick a harmonic oscillator sharply, you impart to it an initial velocity but no initial displacement. For a weakly damped oscillator with mass \(m\), spring constant \(k\). and damping force \(F_{y}=-b v,\) find \(x(t),\) if the total impulse delivered by the kick is \(J_{0}\).

The figure shows a mass \(m_{2}=20.0\) g resting on top of a mass \(m_{1}=20.0 \mathrm{~g}\) which is attached to a spring with \(k=10.0 \mathrm{~N} / \mathrm{m}\) The coefficient of static friction between the two masses is 0.600 . The masses are oscillating with simple harmonic motion on a frictionless surface. What is the maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1} ?\)

With the right choice of parameters, a damped and driven physical pendulum can show chaotic motion, which is sensitively dependent on the initial conditions. Which statement about such a pendulum is true? a) Its long-term behavior can be predicted. b) Its long-term behavior is not predictable. c) Its long-term behavior is like that of a simple pendulum of equivalent length. d) Its long-term behavior is like that of a conical pendulum. e) None of the above is true.

14.3 A mass that can oscillate without friction on a horizontal surface is attached to a horizontal spring that is pulled to the right \(10.0 \mathrm{~cm}\) and is released from rest. The period of oscillation for the mass is \(5.60 \mathrm{~s}\). What is the speed of the mass at \(t=2.50 \mathrm{~s} ?\) a) \(-2.61 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\) b) \(-3.71 \cdot 10^{-2} \mathrm{~m} / \mathrm{s}\) c) \(-3.71 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\) d) \(-2.01 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\)

In a lab, a student measures the unstretched length of a spring as \(11.2 \mathrm{~cm}\). When a 100.0 - g mass is hung from the spring, its length is \(20.7 \mathrm{~cm}\). The mass-spring system is set into oscillatory motion, and the student obscrves that the amplitude of the oscillation decreases by about a factor of 2 after five complete cycles. a) Calculate the period of oscillation for this system, assuming no damping. b) If the student can measure the period to the nearest \(0.05 \mathrm{~s}\). will she be able to detect the difference between the period with no damping and the period with damping?

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