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

The figure shows a mass \(m_{2}=20.0\) g resting on top of a mass \(m_{1}=20.0 \mathrm{~g}\) which is attached to a spring with \(k=10.0 \mathrm{~N} / \mathrm{m}\) The coefficient of static friction between the two masses is 0.600 . The masses are oscillating with simple harmonic motion on a frictionless surface. What is the maximum amplitude the oscillation can have without \(m_{2}\) slipping off \(m_{1} ?\)

A \(3.00-\mathrm{kg}\) mass attached to a spring with \(k=140 . \mathrm{N} / \mathrm{m}\) is oscillating in a vat of oil, which damps the oscillations. a) If the damping constant of the oil is \(b=10.0 \mathrm{~kg} / \mathrm{s}\), how long will it take the amplitude of the oscillations to decrease to \(1.00 \%\) of its original value? b) What should the damping constant be to reduce the amplitude of the oscillations by \(99.0 \%\) in 1.00 s?

A small cylinder of mass m can slide without friction on a shaft that is attached to a turntable, as shown in the figure. The shaft also passes through the coils of a spring with spring constant \(k\), which is attached to the turntable at one end, and to the cylinder at the other end. The equilibrium length of the spring (unstretched and uncompressed) matches the radius of the turntable; thus, when the turntable is not

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 obscrves 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?

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 \(T ?\)

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