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When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of \(12.0 \mathrm{Hz} .\) When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is \(4.00 \mathrm{Hz}\). Find the ratio \(m_{2} / m_{1}\) of the masses.

Short Answer

Expert verified
The ratio \(m_{2}/m_{1}\) is 8.

Step by step solution

01

Understand the Relationship Between Mass and Frequency

In simple harmonic motion, the frequency \( f \) is related to the mass \( m \) and the spring constant \( k \) of the spring by the formula: \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}. \] This implies that \( f^2 = \frac{k}{4\pi^2 m} \).
02

Apply Frequency Formula to Initial Setup

When the object of mass \( m_1 \) is hung on the spring, the frequency is given as \( 12.0 \text{ Hz} \). Substitute this into the equation: \[ (12.0)^2 = \frac{k}{4\pi^2 m_1}. \]
03

Simplify the Initial Frequency Equation

Simplify the equation obtained: \[ 144 = \frac{k}{4\pi^2 m_1}. \] This can be rewritten to express \( k \): \[ k = 144 \times 4\pi^2 m_1. \]
04

Apply Frequency Formula to Combined Masses

When another object \( m_2 \) is added to the spring, the frequency becomes \( 4.00 \text{ Hz} \). Substitute this into the equation: \[ (4.00)^2 = \frac{k}{4\pi^2 (m_1 + m_2)}. \]
05

Simplify the Combined Mass Frequency Equation

Simplify the equation obtained: \[ 16 = \frac{k}{4\pi^2 (m_1 + m_2)}. \] This can be rewritten to express \( k \): \[ k = 16 \times 4\pi^2 (m_1 + m_2). \]
06

Equate Both Expressions for Spring Constant

From Steps 3 and 5, we have two expressions for \( k \). Equate them: \[ 144 \times 4\pi^2 m_1 = 16 \times 4\pi^2 (m_1 + m_2). \]
07

Solve for the Mass Ratio

Cancel the common terms \( 4\pi^2 \) and simplify: \[ 144m_1 = 16(m_1 + m_2). \] Expand and solve for \( m_2 \): \[ 144m_1 = 16m_1 + 16m_2. \] \[ 128m_1 = 16m_2. \] Thus, \[ \frac{m_2}{m_1} = \frac{128}{16} = 8. \]
08

Conclusion

The ratio of the masses \( \frac{m_2}{m_1} = 8 \).

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Key Concepts

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

Mass-Spring System
In the world of physics, the mass-spring system is an elegant model to describe simple harmonic motion. Imagine a spring hanging vertically with a mass attached to the end. This common setup exhibits repeated movements in the form of oscillations. The force exerted by the spring is proportional to the displacement of the mass from its equilibrium position, following Hooke's law, stated as: \[ F = -kx \] where \( F \) is the force applied by the spring, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium.
- **Key Points:**
  • The mass-spring system helps us study oscillatory motion in a simple way.
  • Adding mass to the system will affect its frequency of motion.
  • The spring constant \( k \) is essential in determining the stiffness of the spring and how it behaves when stretched or compressed.
Understanding how these components interact in a mass-spring system is fundamental to exploring more complex harmonic motions.
Frequency
Frequency is a central concept in oscillatory systems, describing how often an action repeats over time. In the context of a mass-spring system, frequency indicates how many complete oscillations or cycles the mass makes per second. It’s measured in Hertz (Hz), where one Hertz equals one cycle per second.In simple harmonic motion, the frequency \( f \) is directly related to the mass \( m \) attached to the spring, as well as the spring constant \( k \), by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
- **Key Insights:**
  • Frequency depends on both the mass and the spring constant.
  • Increasing the mass usually lowers the frequency since the system becomes slower.
  • High frequency implies quick oscillations, while low frequency suggests slower movement.
This relationship is crucial because it helps us predict how changes to mass or the spring constant will affect the motion of the system.
Harmonic Oscillation
Harmonic oscillation refers to the repetitive, periodic motion often evidenced in a mass-spring system. This type of motion is characterized by its sinusoidal pattern, which is smooth and consistent over time. Such motion abides by principles of energy conservation, oscillating due to a balance between kinetic and potential energy. This energy transfer leads to the elegant back-and-forth movement seen in many everyday oscillating systems, following the formula:
\[ x(t) = A \cos(\omega t + \phi) \]
Here, \( x(t) \) describes the position over time, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.- **Key Characteristics:**
  • Harmonic oscillations have fixed frequencies and are predictable.
  • The motion remains stable as long as no external forces interfere with the system.
  • This oscillation is termed 'simple' when driven purely by a linear restoring force like a spring.
Understanding harmonic oscillation enhances our comprehension of various natural and mechanical systems where such motion is prevalent.

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Most popular questions from this chapter

A heavy-duty stapling gun uses a \(0.140-\mathrm{kg}\) metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

A 1.1-kg object is suspended from a vertical spring whose spring constant is \(120 \mathrm{N} / \mathrm{m}\). (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of \(0.20 \mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amplitude of \(6.3 \times 10^{-7}\) m and a maximum speed of \(2.9 \times 10^{-3} \mathrm{m} / \mathrm{s}\) (a) What is the frequency (in Hz) of the eardrum’s vibration? (b) What is the maximum acceleration of the eardrum?

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{m}\), released from rest, and subsequently oscillate back and forth with an angular frequency of 11.3 rad/s. What is the speed of the object at the instant when the spring is stretched hy \(0.048 \mathrm{m}\) relative to its \(\mathrm{un}\) strained length?

A \(1.00 \times 10^{-2}\) -kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is \(124 \mathrm{N} / \mathrm{m} .\) The block is shoved parallel to the spring axis and is given an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

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