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Object \(A\) is four times heavier than object B. Fach object is attached to a spring, and the springs have equal spring constants. The two objects are then pulled from their equilibrium positions and released from rest. What is the ratio of the periods of the two oscillators if the amplitude of \(A\) is half that of \(B ?\) a) \(T_{A}: T_{\mathrm{g}}=1: 4\) c) \(T_{A}: T_{\mathrm{B}}=2\) : b) \(T_{A}: T_{B}=4: 1\) d) \(T_{A}: T_{B}=1: 2\)

Short Answer

Expert verified
Answer: The ratio of the periods is 2 : 1.

Step by step solution

01

Identify the formula for the period of a simple harmonic oscillator

For a simple harmonic oscillator, the period (T) is given by the formula: \(T = 2\pi\sqrt{\frac{m}{k}}\), where \(m\) is the mass of the object, and \(k\) is the spring constant.
02

Set up the variables for objects A and B

Let \(m_A = 4m_B\) be the mass of object A, and \(m_B\) be the mass of object B since object A is four times heavier than object B. The spring constants are equal, so let \(k_A = k_B = k\).
03

Calculate the periods for objects A and B

Using the period formula, we will calculate the periods for objects A and B: \(T_A = 2\pi\sqrt{\frac{m_A}{k}} = 2\pi\sqrt{\frac{4m_B}{k}}\) \(T_B = 2\pi\sqrt{\frac{m_B}{k}}\)
04

Determine the ratio of the periods

Now we will find the ratio of the periods: \(\frac{T_A}{T_B} = \frac{2\pi\sqrt{\frac{4m_B}{k}}}{2\pi\sqrt{\frac{m_B}{k}}}\) Cancel out the common factors and simplify: \(\frac{T_A}{T_B} = \frac{\sqrt{4m_B}}{\sqrt{m_B}} = \frac{2\sqrt{m_B}}{\sqrt{m_B}} = 2\) So the ratio of the periods is \(T_A : T_B = 2 : 1\), which corresponds to answer choice (c).

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Key Concepts

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

Oscillation Period
The oscillation period, often denoted as T, is a fundamental concept when exploring simple harmonic motion. It describes the time it takes for an object to complete one full cycle of motion – from its starting position, through its maximum displacement, to its opposite extreme, and then back again to the initial position. For a mass-spring system, the period T is determined by using the formula:

\[\begin{equation}T = 2\text{\pi}\sqrt{\frac{m}{k}}\text{\end{equation}\]}
Where:
  • m represents the mass of the object attached to the spring
  • k is the spring constant, which is a measure of the stiffness of the spring
Importantly, the period depends on the square root of the mass and is inversely proportional to the square root of the spring constant, meaning if the mass increases, the period will also increase, and if the spring constant increases, the period decreases. This concept is crucial for understanding how different factors affect the motion of oscillating systems.
Spring Constant
Another core component of simple harmonic motion is the spring constant, denoted by the symbol k. This constant is a measure of the stiffness of a spring and dictates how resistant it is to being compressed or stretched. Mathematically, it's defined by Hooke's Law, which states that the force F exerted by a spring is directly proportional to the displacement x that is applied to it:

\[F = -kx\]
In this context, the negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. The higher the spring constant, the more force is needed to stretch or compress it by a given amount. This characteristic directly influences the oscillation period of mass-spring systems, as mentioned in the previous section.
Mass-Spring System
When diving into the dynamics of a mass-spring system, one must carefully consider both the mass of the object and the spring constant of the spring involved. A mass-spring system consists of a mass (m) that is attached to a spring with a certain spring constant (k). This type of system can undergo simple harmonic motion when displaced from its equilibrium position.

The motion of such a system is predictable and repeatable, characterized by a back-and-forth oscillation about the equilibrium position. A fundamental property of the mass-spring system is that its motion is determined solely by m and k, regardless of the amplitude of oscillation – which means how far the mass is pulled or pushed from the equilibrium position does not affect the system's period, a fact reflected in the period formula and the exercise. This concept paves the way for a wide range of applications, including timekeeping in watches and seismology, where the principles governing mass-spring systems are utilized to interpret the vibrations of the Earth.

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Most popular questions from this chapter

A physical pendulum consists of a uniform rod of mass \(M\) and length \(L\) The pendulum is pivoted at a point that is a distance \(x\) from the center of the rod, so the period for oscillation of the pendulum depends on \(x: T(x)\). a) What value of \(x\) gives the maximum value for \(T ?\) b) What value of \(x\) gives the minimum value for \(T ?\)

A mass of \(0.404 \mathrm{~kg}\) is attached to a spring with a spring constant of \(206.9 \mathrm{~N} / \mathrm{m}\). Its oscillation is damped. with damping constant \(b=14.5 \mathrm{~kg} / \mathrm{s}\). What is the frequency of this damped oscillation?

An \(80.0-\mathrm{kg}\) bungee jumper is enjoying an afternoon of jumps. The jumper's first oscillation has an amplitude of \(10.0 \mathrm{~m}\) and a period of \(5.00 \mathrm{~s}\). Treating the bungee cord as spring with no damping, calculate each of the following: a) the spring constant of the bungee cord. b) the bungee jumper's maximum speed during the ascillation, and c) the time for the amplitude to decrease to \(2.00 \mathrm{~m}\) (with air resistance providing the damping of the oscillations at \(7.50 \mathrm{~kg} / \mathrm{s})\)

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 onedimensional, 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 Foucault pendulum displayed in a museum is typically quite long, making the effect easier to see. Consider a Foucault pendulum of length \(15 \mathrm{~m}\) with a 110 -kg brass bob. It is set to swing with an amplitude of \(3.5^{\circ}\) a) What is the period of the pendulum? b) What is the maximum kinetic energy of the pendulum? c) What is the maximum speed of the pendulum?

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