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A \(2.00-\mathrm{kg}\) object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is \(5.00 \mathrm{~N} / \mathrm{m}\). The object is displaced \(3.00 \mathrm{~m}\) to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object \(3.50 \mathrm{~s}\) after it is released? (b) How many times does the object oscillate in \(3.50 \mathrm{~s}\) ?

Short Answer

Expert verified
The force acting on the object 3.50 s after it is released is 15.00 N to the left. The object oscillates approximately 1.25 times in 3.50 s.

Step by step solution

01

Apply Hooke's Law

According to Hooke's law, the force exerted by a spring is given by \( F = -kx \), where \( k \) is the force constant of the spring and \( x \) is the displacement of the mass from the equilibrium position. Here, since the object is released from 3.00 m to the right of the equilibrium position, the displacement \( x = -3.00 m \). The negative sign here indicates that the force is directed to the left. The force constant \( k = 5.00 N/m \). So, compute force as \( F = -kx = - (5.00 N/m) (-3.00 m) = 15.00 N \)
02

Calculate period of oscillation

The period of oscillation of a spring, denoted as \( T \), is given by the formula \( T = 2 \pi \sqrt{m/k} \), where \( m \) is the mass of the object and \( k \) is the spring constant. Plugging in the values \( m = 2.00 kg \) and \( k = 5.00 N/m \) gives \( T = 2 \pi \sqrt{2.00 kg / 5.00 N/m} \)
03

Calculate time per oscillation

To determine how many times the object oscillates in 3.50 s, you need to know how much time it takes for one oscillation. This is given by the period. Divide the total time by period T computed in the previous step to find the number of oscillations.

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Key Concepts

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

Hooke's Law
Hooke's Law is essential in understanding how springs behave in simple harmonic motion. The law describes how the force exerted by a spring relates to its displacement from its equilibrium position. Hooke's Law can be expressed mathematically by the equation:
  • \( F = -kx \)
Here:
  • \( F \) is the force exerted by the spring in newtons (N).
  • \( k \) is the spring constant, indicating the stiffness of the spring, measured in N/m.
  • \( x \) is the displacement from the spring's equilibrium position in meters (m).

In the given exercise, when a 2.00 kg object is displaced 3.00 m from its equilibrium position, Hooke's Law provides the framework to determine the force acting upon it. Importantly, the negative sign in the equation indicates the direction of the force is opposite to the direction of displacement. If the object is moved to the right, the spring's force pulls it back to the left. This characteristic of oppositional force is fundamental to the nature of oscillatory motion.
Oscillation
Oscillation refers to the repetitive back-and-forth movement of an object, such as a mass attached to a spring. Understanding this concept is crucial to solving such problems. When the mass is displaced from its equilibrium position and released, it will continuously move back and forth due to the restoring force provided by the spring.

The complete movement from the starting point to the opposite side and back defines one full oscillation. The time taken for one complete oscillation is known as the period (\( T \)), calculated using:
  • \( T = 2 \pi \sqrt{ \frac{m}{k} } \)

This formula involves:
  • \( m \) – mass of the object (kg).
  • \( k \) – spring constant (N/m).

In practical terms, after determining the period, we can compute how often the object oscillates within a given time frame. For example, if you know \( T \) and you have a time duration, you can easily calculate the number of oscillations by dividing the total time by the period. This logical process helps to understand the inherent rhythm of harmonic systems.
Spring Constant
The spring constant \( k \) is a measure of a spring's stiffness. It's a crucial factor that influences how a spring will behave under force.
  • A higher \( k \) value means a stiffer spring, which requires more force to stretch or compress a given distance.
  • A lower \( k \) value indicates a more flexible spring, which is easier to stretch or compress.

In the context of simple harmonic motion, the spring constant affects both the force exerted by the spring and the period of oscillation. For instance, in our exercise with a spring constant of 5.00 N/m, we anticipate a certain degree of stiffness:
  • This allows us to calculate the force exerted when the object is displaced and the period of oscillation based on that stiffness.
  • The accuracy and relevancy of \( k \) in these calculations reflects the spring's predictable behavior.

Understanding the spring constant gives us insight into not only the immediate scenario but also broader applications, like shock absorbers in vehicles or measuring devices like scales. Thus, \( k \) serves as a key parameter in the dynamics of a spring-mass system.

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Most popular questions from this chapter

A "seconds" pendulum is one that moves through its equilibrium position once each second. (The period of the pendulum is \(2.000 \mathrm{~s}\).) The length of a seconds pendulum is \(0.9927 \mathrm{~m}\) at Tokyo and \(0.9942 \mathrm{~m}\) at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?

A \(3.00-\mathrm{kg}\) object is fastened to a light spring, with the intervening cord passing over a pulley (Fig. P13.67). The pulley is frictionless, and its inertia may be neglected. The object is released from rest when the spring is unstretched. If the object drops \(10.0 \mathrm{~cm}\) before stopping, find (a) the spring constant of the spring and (b) the speed of the object when it is \(5.00 \mathrm{~cm}\) below its starting point.

The position of an object connected to a spring varies with time according to the expression \(x=\) \((5.2 \mathrm{~cm}) \sin (8.0 \pi t)\). Find (a) the period of this motion, (b) the frequency of the motion, (c) the amplitude of the motion, and (d) the first time after \(t=0\) that the object reaches the position \(x=2.6 \mathrm{~cm}\).

A horizontal spring attached to a wall has a force constant of \(k=850 \mathrm{~N} / \mathrm{m}\). A block of mass \(m=1.00 \mathrm{~kg}\) is attached to the spring and rests on a frictionless, horizontal surface as in Figure P13.21. (a) The block is pulled to a position \(x_{i}=6.00 \mathrm{~cm}\) from equilibrium and released. Find the potential energy stored in the spring when the block is \(6.00 \mathrm{~cm}\) from equilibrium. (b) Find the speed of the block as it passes through the equilibrium position. (c) What is the speed of the block when it is at a position \(x_{i} / 2=3.00 \mathrm{~cm}\) ?

A \(10.0-\mathrm{g}\) bullet is fired into, and embeds itself in, a \(2.00-\mathrm{kg}\) block attached to a spring with a force constant of \(19.6 \mathrm{~N} / \mathrm{m}\) and whose mass is negligible. How far is the spring compressed if the bullet has a speed of \(300 \mathrm{~m} / \mathrm{s}\) just before it strikes the block and the block

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