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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 simple pendulum with a length of \(2.23 \mathrm{m}\) and a mass of \(6.74 \mathrm{kg}\) is given an initial speed of \(2.06 \mathrm{m} / \mathrm{s}\) at its equilibrium position. Assume it undergoes simple harmonic motion, and determine its (a) period, (b) total energy, and (c) maximum angular displacement.

One end of a light spring with force constant \(100 \mathrm{N} / \mathrm{m}\) is attached to a vertical wall. A light string is tied to the other end of the horizontal spring. The string changes from horizontal to vertical as it passes over a solid pulley of diameter \(4.00 \mathrm{cm} .\) The pulley is free to turn on a fixed smooth axle. The vertical section of the string supports a \(200-\mathrm{g}\) object. The string does not slip at its contact with the pulley. Find the frequency of oscillation of the object if the mass of the pulley is (a) negligible, (b) \(250 \mathrm{g}\), and (c) \(750 \mathrm{g}\)

The initial position, velocity, and acceleration of an object moving in simple harmonic motion are \(x_{i}, v_{i},\) and \(a_{i} ;\) the angular frequency of oscillation is \(\omega .\) (a) Show that the position and velocity of the object for all time can be written as,$$\begin{array}{l}x(t)=x_{i} \cos \omega t+\left(\frac{v_{i}}{\omega}\right) \sin \omega t \\\v(t)=-x_{i} \omega \sin \omega t+v_{i}\cos \omega t\end{array}$$ (b) If the amplitude of the motion is \(A\), show that,$$v^{2}-a x=v_{i}^{2}-a_{i} x_{i}=\omega^{2} A^{2}$$.

A 50.0 -g object connected to a spring with a force constant of \(35.0 \mathrm{N} / \mathrm{m}\) oscillates on a horizontal, frictionless surface with an amplitude of \(4.00 \mathrm{cm} .\) Find (a) the total energy of the system and (b) the speed of the object when the position is \(1.00 \mathrm{cm} .\) Find \((\mathrm{c})\) the kinetic energy and \((\mathrm{d})\) the potential energy when the position is \(3.00 \mathrm{cm} .\)

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?

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