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A block-spring system oscillates with an amplitude of \(3.50 \mathrm{cm} .\) If the spring constant is \(250 \mathrm{N} / \mathrm{m}\) and the mass of the block is \(0.500 \mathrm{kg},\) determine (a) the mechanical energy of the system, (b) the maximum speed of the block, and (c) the maximum acceleration.

Short Answer

Expert verified
The mechanical energy of the system is 0.30625 J, the maximum speed of the block is 0.7826 m/s and the maximum acceleration of the block is 17.52 m/s^2.

Step by step solution

01

Calculate Mechanical Energy

The formula for mechanical energy in an oscillator is \(0.5*k*A^2\). Substituting \(k=250 N/m\) and \(A=0.035 m\) (because 3.5 cm = 0.035 m), the mechanical energy will be \(0.5*250*0.035^2 = 0.30625 J\).
02

Calculate Maximum Speed

The maximum speed \(v_{max}= \omega \times A\) where \(\omega= \sqrt{\frac{k}{m}}\). First find \(\omega= \sqrt{\frac{250}{0.5}} = 22.36 rad/s\). Afterwards, find \(v_{max}= 22.36 * 0.035 = 0.7826 m/s\).
03

Calculate Maximum Acceleration

Max acceleration is given by \( \omega^2 * A = 22.36^2 * 0.035 = 17.52 m/s^2\).

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Key Concepts

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

Mechanical Energy of Oscillators
The mechanical energy of an oscillator is a measure of the total energy possessed by the system due to its motion and position. In the case of a mass attached to a spring (block-spring system), the energy alternates between kinetic energy, when the mass is in motion, and potential energy, when the spring is compressed or elongated. The mechanical energy in an ideal oscillator is constant, given there are no external forces such as friction causing energy loss.

Using the formula for mechanical energy, which is \(0.5 \times k \times A^2\), we can determine the total energy in the system throughout its motion. Here, \(k\) represents the spring constant, which describes the stiffness of the spring, and \(A\) is the amplitude of oscillation, the maximum extent of displacement from the equilibrium position. In context, given \(k = 250 \text{N/m}\) and \(A = 0.035 \text{m}\) for our block-spring system, we calculated the mechanical energy to be 0.30625 joules.

Understanding mechanical energy in oscillators is crucial because it remains conserved in a frictionless environment, ensuring that the motion continues indefinitely. This principle applies widely in many mechanical systems and is foundational in understanding harmonic motion.
Maximum Speed in Oscillatory Motion
The maximum speed in oscillatory motion is a critical point at which the moving object, like the block in our spring system, reaches its highest velocity. It occurs as the oscillator passes through the equilibrium point, where all the mechanical energy is temporarily converted into kinetic energy.

To calculate this maximum speed (\(v_{max}\)), we utilize the relationship \(v_{max} = \omega \times A\), involving the angular frequency \(\omega\) and the amplitude \(A\) of the oscillator. Angular frequency (\(\omega\)) is determined by the formula \(\omega = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant, and \(m\) is the mass of the oscillator. For our spring-block scenario with an amplitude of 0.035 meters, and \(\omega\) calculated to be 22.36 rad/s, the maximum speed produced is 0.7826 meters per second.

Grasping the concept of maximum speed is essential for predicting the behavior of oscillatory systems and designing mechanisms like suspensions or seismic isolators where managing the point of highest kinetic energy is fundamental.
Maximum Acceleration in Simple Harmonic Motion
When discussing oscillations, acceleration is another pivotal aspect to consider, particularly the maximum acceleration which occurs at the points of maximum displacement from the equilibrium - in other words, at the amplitude points. The formula for the maximum acceleration is given by \(a_{max} = \omega^2 \times A\), where \(\omega\) is again our angular frequency, and \(A\) signifies the amplitude.

In our given exercise, the maximum acceleration is achieved by squaring the previously calculated angular frequency, 22.36 rad/s, and multiplying by the amplitude of 0.035 meters. This calculation results in a maximum acceleration of 17.52 meters per second squared. This value indicates how quickly the velocity of the oscillator changes at its most extreme points.

Understanding the concept of maximum acceleration in simple harmonic motion allows for insights into the dynamic forces at play within oscillatory systems. For instance, engineers consider these forces when designing shock absorbers and crafting earthquake-resistant structures to ensure safety and stability under conditions that induce oscillatory motion.

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Most popular questions from this chapter

A \(2.00-\mathrm{kg}\) object attached to a spring moves without friction and is driven by an external force given by \(F=\) \((3.00 \mathrm{N}) \sin (2 \pi t) .\) If the force constant of the spring is 20.0 N/m, determine (a) the period and (b) the amplitude of the motion.

A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2 s.) The length of a seconds pendulum is \(0.9927 \mathrm{m}\) at Tokyo, Japan and \(0.9942 \mathrm{m}\) at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?

(a) A hanging spring stretches by \(35.0 \mathrm{cm}\) when an object of mass \(450 \mathrm{g}\) is hung on it at rest. In this situation, we define its position as \(x=0 .\) The object is pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. What is its position \(x\) at a time 84.4 s later? (b) What If? A hanging spring stretches by \(35.5 \mathrm{cm}\) when an object of mass \(440 \mathrm{g}\) is hung on it at rest. We define this new position as \(x=0 .\) This object is also pulled down an additional \(18.0 \mathrm{cm}\) and released from rest to oscillate without friction. Find its position 84.4 s later. (c) Why are the answers to (a) and (b) different by such a large percentage when the data are so similar? Does this circumstance reveal a fundamental difficulty in calculating the future? (d) Find the distance traveled by the vibrating object in part (a). (e) Find the distance traveled by the object in part (b).

A particle executes simple harmonic motion with an amplitude of \(3.00 \mathrm{cm} .\) At what position does its speed equal half its maximum speed?

A simple pendulum is \(5.00 \mathrm{m}\) long. (a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at \(5.00 \mathrm{m} / \mathrm{s}^{2} ?\) (b) What is its period if the elevator is accelerating downward at \(5.00 \mathrm{m} / \mathrm{s}^{2} ?\) (c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at \(5.00 \mathrm{m} / \mathrm{s}^{2} ?\)

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