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 mass, \(M=1.6 \mathrm{~kg}\), is attached to a wall by a spring with \(k=578 \mathrm{~N} / \mathrm{m}\). The mass slides on a frictionless floor. The spring and mass are immersed in a fluid with a damping constant of \(6.4 \mathrm{~kg} / \mathrm{s}\). A horizontal force, \(\mathrm{F}(\mathrm{t})=\mathrm{F}_{\mathrm{d}} \cos \left(\omega_{\mathrm{d}} \mathrm{t}\right)\) where \(F_{d}=52 \mathrm{~N},\) is applied to the mass through a knob, caus. ing the mass to oscillate back and forth. Neglect the mass of the spring and of the knob and rod. At approximately what frequency will the amplitude of the mass's oscillation be greatest, and what is the maximum amplitude? If the driving frequency is reduced slightly (but the driving amplitude remains the same), at what frecuency will the amplitude of the mass's ascillation be half of the maximum amplitude?

The period of oscillation of an object in a frictionless tunnel running through the center of the Moon is \(T=2 \pi / \omega_{0}\) \(=6485 \mathrm{~s}\), as shown in Fxample 142 . What is the period of oscillation of an object in a similar tunnel through the Earth \(\left(R_{\mathrm{I}}=6.37 \cdot 10^{6} \mathrm{~m} ; R_{\mathrm{M}}=1.74 \cdot 10^{6} \mathrm{~m} ; M_{\mathrm{E}}=5.98 \cdot 10^{24} \mathrm{~kg}\right.\) \(\left.M_{u}=7.36 \cdot 10^{22} \mathbf{k g}\right) ?\)

A 2.0 -kg mass attached to a spring is displaced \(8.0 \mathrm{~cm}\) from the equilibrium position. It is released and then oscillates with a frequency of \(4.0 \mathrm{~Hz}\) a) What is the energy of the motion when the mass passes through the equilibrium position? b) What is the speed of the mass when it is \(20 \mathrm{~cm}\) from the equilibrium position?

A mass \(m=5.00 \mathrm{~kg}\) is suspended from a spring and oscillates according to the equation of motion \(x(t)=0.5 \cos (5 t+\pi / 4) .\) What is the spring constant?

A 3.0 -kg mass is vibrating on a spring. It has a resonant angular speed of \(2.4 \mathrm{rad} / \mathrm{s}\) and a damping angular speed of \(0.14 \mathrm{rad} / \mathrm{s}\). If the driving force is \(2.0 \mathrm{~N}\), find the maximum amplitude if the driving angular speed is (a) \(1.2 \mathrm{rad} / \mathrm{s},\) (b) \(2.4 \mathrm{rad} / \mathrm{s}\), and \((\) c) \(4.8 \mathrm{rad} / \mathrm{s}\).

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