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A horizontal object-spring system oscillates with an amplitude of \(3.5 \mathrm{~cm}\) on a frictionless surface. If the spring constant is \(250 \mathrm{~N} / \mathrm{m}\) and the object has a mass of \(0.50 \mathrm{~kg}\), determine (a) the mechanical energy of the system, (b) the maximum speed of the object, and (c) the maximum acceleration of the object.

Short Answer

Expert verified
The total mechanical energy of the system is approximately 3.06 J. The maximum speed of the object is approximately 7.85 m/s. The maximum acceleration of the object is approximately 17.5 m/s^2.

Step by step solution

01

Calculate the Angular Frequency

The angular frequency \(ω\) is given by the formula \(\sqrt{k/m}\). Substituting \(k = 250 N/m\) and \(m = 0.50 kg\) into the formula gives \(ω = √(250/0.50) = √500 rad/s\).
02

Calculate the Total Mechanical Energy

The total mechanical energy \(E\) in the system is given by \(E = 0.5*m*(ωA)^2\). Substitute the values \(m = 0.50 kg\), \(A = 0.035 m\) and \(ω = √500 rad/s\) into the formula. This gives \(E = 0.5*0.50*(√500*0.035)^2 ≈ 3.06 J\) for the total mechanical energy.
03

Calculate the Maximum Speed

The maximum speed \(v\) of the object is given by \(v = ωA\). Substitute the values \(ω = √500 rad/s\) and \(A = 0.035 m\) into the formula. This gives \(v = √500*0.035 ≈ 7.85 m/s\)
04

Calculate the Maximum Acceleration

The maximum acceleration \(a\) of the object is given by \(a = ω^2*A\). Substitute the values \(ω = √500 rad/s\) and \(A = 0.035 m\) into the formula. This gives \(a = 500*0.035 ≈ 17.5 m/s^2\)

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Key Concepts

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

Spring Constant
In the context of harmonic motion, the spring constant, often represented by the letter "k," is a crucial factor. It measures the stiffness of a spring. The larger the spring constant, the stiffer the spring, and the more force is needed to stretch or compress it by a certain amount. The spring constant is measured in newtons per meter (N/m).
The formula to find the force that a spring exerts is given by Hooke's Law:
  • \( F = -kx \)
where:
  • \( F \) is the force in newtons.
  • \( k \) is the spring constant in newtons per meter.
  • \( x \) is the displacement from the spring's equilibrium position in meters.
This fundamental concept is vital for solving various problems in physics that involve oscillating systems.
Angular Frequency
Angular frequency is a term often used to describe how rapidly an object moves through its cycle of motion. It is closely related to linear frequencies and represents how fast the oscillations occur per unit time. The mathematical symbol for angular frequency is \( \omega \) (omega) and its unit is radians per second (rad/s).
The formula to calculate angular frequency in a spring-mass system is:
  • \( \omega = \sqrt{\frac{k}{m}} \)
where:
  • \( k \) is the spring constant in N/m.
  • \( m \) is the mass of the oscillating object in kilograms.
This means that the angular frequency increases with a stiffer spring or a lighter mass.
Understanding angular frequency is essential because it allows us to determine related properties such as the system's mechanical energy and the maximum speeds and accelerations.
Mechanical Energy
Mechanical energy in a harmonic oscillator involves both kinetic and potential energy. In the context of a horizontal spring system, it represents the total energetic disposition in the form of motion and stored energy. The total mechanical energy for such a system remains constant if no non-conservative forces like friction are present.
The formula to calculate mechanical energy \( E \) in a spring-mass system is given by:
  • \( E = \frac{1}{2} k A^2 \)
where:
  • \( A \) is the amplitude of motion in meters.
  • \( k \) is the spring constant in N/m.
In our problem, the mechanical energy can also be expressed as \( E = \frac{1}{2} m (\omega A)^2 \), illustrating the energy contained as the oscillator changes states from kinetic to potential energy.Understanding the concept of mechanical energy helps in grasping how energy is conserved within isolated systems.
Maximum Speed
The maximum speed for an object in harmonic motion is significant since it corresponds to the point where kinetic energy is at its maximum and potential energy is zero.
  • The formula to find maximum speed \( v_{max} \) is:
  • \( v_{max} = \omega A \)
where:
  • \( \omega \) is the angular frequency in rad/s.
  • \( A \) is the amplitude in meters.
The maximum speed occurs when the object passes through the equilibrium position during its oscillation. At this moment, the energy transformation between kinetic and potential forms is at its peak.
Recognizing the concept of maximum speed allows a deeper understanding of how oscillatory systems store and release energy over their motion cycles.

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Most popular questions from this chapter

A vertical spring stretches \(3.9 \mathrm{~cm}\) when a \(10-\mathrm{g}\) object is hung from it. The object is replaced with a block of mass \(25 \mathrm{~g}\) that oscillates up and down in simple harmonic motion. Calculate the period of motion.

A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes \(40.0\) vibrations in \(30.0 \mathrm{~s}\). Also, a given maximum travels \(425 \mathrm{~cm}\) along the rope in \(10.0 \mathrm{~s}\). What is the wavelength?

A \(0.250-\mathrm{kg}\) block resting on a frictionless, horizontal surface is attached to a spring having force constant \(\quad 83.8 \mathrm{~N} / \mathrm{m}\) as in Figure P13.16. A horizontal force \(\overrightarrow{\mathbf{F}}\) causes the spring to stretch a distance of \(5.46 \mathrm{~cm}\) from its equilibrium position. (a) Find the value of \(F\). (b) What is the total energy stored in the system when the spring is stretched? (c) Find the magnitude of the acceleration of the block immediately after the applied force is removed. (d) Find the speed of the block when it first reaches the equilibrium position. (e) If the surface is not frictionless but the block still reaches the equilibrium position, how would your answer to part (d) change? (f) What other information would you need to know to find the actual answer to part (d) in this case?

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy \(E=\) \(47.0 \mathrm{~J}\) and a maximum displacement from equilibrium of \(0.240 \mathrm{~m}\). (a) What is the spring constant? (b) What is the kinetic energy of the system at the equilibrium point? (c) If the maximum speed of the block is \(3.45 \mathrm{~m} / \mathrm{s}\), what is its mass? (d) What is the speed of the block when its displacement is \(0.160 \mathrm{~m}\) ? (e) Find the kinetic energy of the block at \(x=0.160 \mathrm{~m}\). (f) Find the potential energy stored in the spring when \(x=0.160 \mathrm{~m}\). (g) Suppose the same system is released from rest at \(x=0.240 \mathrm{~m}\) on a rough surface so that it loses \(14.0 \mathrm{~J}\) by the time it reaches its first turning point (after passing equilibrium at \(x=0\) ). What is its position at that instant?

A spring in a toy gun has a spring constant of \(9.80 \mathrm{~N} / \mathrm{m}\) and can be compressed \(20.0 \mathrm{~cm}\) beyond the equilibrium position. A \(1.00-\mathrm{g}\) pellet resting against the spring is propelled forward when the spring is released. (a) Find the muzzle speed of the pellet. (b) If the pellet is fired horizontally from a height of \(1.00 \mathrm{~m}\) above the floor, what is its range?

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