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Identical blocks oscillate on the end of a vertical spring, one on Earth and one on the Moon. Where is the period of the oscillations greater? a) on Earth b) on the Moon c) same on both Earth and Moon d) cannot be determined from the information given.

Short Answer

Expert verified
Answer: c) same on both Earth and Moon

Step by step solution

01

Determine the formula for the period of oscillations

To find the period of oscillations, we will use the formula for the period of a mass-spring system which is given by: T = 2π√(m/k) Where T is the period, m is the mass of the object, and k is the spring constant.
02

Identify the key difference between Earth and the Moon

The main difference between Earth and the Moon that affects the period of oscillations is gravity. The gravitational force on the Moon is about 1/6th of that on Earth.
03

Determine the effect of gravity on the mass-spring system

However, if we look closely at the formula for the period of oscillations, we can see that gravity has no direct impact in it. The period of oscillations only depends on the mass of the object and the spring constant. Gravity does not have any effect on these variables.
04

Conclude the answer based on the analysis

Since gravity doesn't have any direct impact on the period of oscillations, we can conclude that the period will be the same both on Earth and on the Moon. Therefore, the correct answer is: c) same on both Earth and Moon

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Key Concepts

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

Mass-Spring System
Understanding the mass-spring system is critical to grasping the concept of periodic motion in physics. This system comprises a spring with a spring constant, often denoted as 'k', and a mass, 'm', that is attached to the spring.

The physics behind the system lies in Hooke's Law, which states that the force exerted by the spring is directly proportional to the displacement from its equilibrium position. This means that the more you stretch or compress the spring, the greater the restoring force that tries to bring it back to its original state.

Now, when the mass is set into oscillation, that is, periodically moving back and forth, the mass-spring system undergoes a type of motion known as simple harmonic motion. In an ideal scenario, where there's no friction or air resistance, the system would keep oscillating indefinitely.

The oscillation's period, which is the time taken for one complete cycle, is a critical aspect of the mass-spring system. It is independent of the amplitude of oscillations and is calculated using the formula: \[ T = 2\pi\sqrt{\frac{m}{k}} \] where 'T' denotes the period.

This fundamental understanding aids in predicting how altering the mass or the spring constant would affect the system’s behavior, a concept that has far-reaching implications in fields ranging from mechanical engineering to seismology.
Gravitational Force
Gravitational force is a fundamental natural force that governs the motion of bodies under the influence of mass. It's the attractive force that every object with mass exerts on every other object with mass.

The importance of this force spans from our daily experiences, like keeping our feet firmly on the ground, to astronomical scales, where it governs the motion of planets, stars, and galaxies. In the context of physics problems, especially those involving oscillations, one might assume that gravity would play an integral role.

However, it's key to understand that while gravitational force is omnipresent, its impact on certain systems can vary. For the period of oscillation of a mass-spring system in particular, gravity does not directly come into play. This is because the formula for the period of the system's oscillation, as we've previously addressed, only incorporates the mass of the object and the spring constant.

Nonetheless, gravity is crucial when considering the weight of the object, as in other types of pendulum or freefall problems. It's also essential to note that gravity's effects differ across celestial bodies, influencing the weight of the objects but not their mass.
Spring Constant
The spring constant, symbolized as 'k', is essential in describing the stiffness of a spring. It is a measure of the resistance to being compressed or stretched - the higher the spring constant, the stiffer the spring is, and the more force is required to deform it.

In the context of oscillations within a mass-spring system, the spring constant is a cornerstone of the formula used to calculate the period: \[ T = 2\pi\sqrt{\frac{m}{k}} \]. This formula showcases that the period is inversely proportional to the square root of the spring constant.

It’s fascinating that the spring constant is independent of gravitational effects. This property allows the period of oscillation to remain constant for a mass-spring system, even when the location changes from Earth to the Moon.

The spring constant doesn't only come up in classical mechanics and engineering design but also in the design of everyday objects like mattresses, vehicle suspensions, and even in biomedical engineering applications such as artificial limbs. Understanding 'k' can reveal much about the behavior of a spring and the systems in which they are integral components.

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Most popular questions from this chapter

The motion of a planet in a circular orbit about a star obeys the equations of simple harmonic motion. If the orbit is observed edge-on, so the planet's motion appears to be one-dimensional, the analogy is quite direct: The motion of the planet looks just like the motion of an object on a spring. a) Use Kepler's Third Law of planetary motion to determine the "spring constant" for a planet in circular orbit around a star with period \(T\). b) When the planet is at the extremes of its motion observed edge-on, the analogous "spring" is extended to its largest displacement. Using the "spring" analogy, determine the orbital velocity of the planet.

A vertical spring with spring constant \(23.73 \mathrm{~N} / \mathrm{m}\) is hanging from a ceiling. A small object with mass \(1.103 \mathrm{~kg}\) is attached to the lower end of the spring, and the spring stretches to its equilibrium length. The object is then pulled down and released. The speed of the object when it is \(4.985 \mathrm{~cm}\) from the equilibrium position is \(0.4585 \mathrm{~m} / \mathrm{s}\). How far down was the object pulled?

The spring constant for a spring-mass system undergoing simple harmonic motion is doubled. If the total energy remains unchanged, what will happen to the maximum amplitude of the oscillation? Assume that the system is underdamped. a) It will remain unchanged. b) It will be multiplied by 2 . c) It will be multiplied by \(\frac{1}{2}\) d) It will be multiplied by \(1 / \sqrt{2}\).

An object in simple harmonic motion is isochronous, meaning that the period of its oscillations is independent of their amplitude. (Contrary to a common assertion, the operation of a pendulum clock is not based on this principle. A pendulum clock operates at fixed, finite amplitude. The gearing of the clock compensates for the anharmonicity of the pendulum. Consider an oscillator of mass \(m\) in one-dimensional motion, with a restoring force \(F(x)=-c x^{3},\) where \(x\) is the displacement from equilibrium and \(c\) is a constant with appropriate units. The motion of this oscillator is periodic but not isochronous. a) Write an expression for the period of the undamped oscillations of this oscillator. If your expression involves an integral, it should be a definite integral. You do not need to evaluate the expression. b) Using the expression of part (a), determine the dependence of the period of oscillation on the amplitude. c) Generalize the results of parts (a) and (b) to an oscillator of mass \(m\) in one-dimensional motion with a restoring force corresponding to the potential energy \(U(x)=\gamma|x|^{\alpha} / \alpha,\) where \(\alpha\) is any positive value and \(\gamma\) is a constant.

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 observes 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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