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When a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 Hz. What will the frequency be if 0.220 kg are added to the original mass, and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

Short Answer

Expert verified
Adding 0.220 kg, the frequency is approximately 1.17 Hz. Subtracting 0.220 kg, the frequency is approximately 1.56 Hz.

Step by step solution

01

Understanding the Relationship

The frequency of an oscillating mass-spring system is given by \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass attached to the spring. We need to find how this frequency changes when the mass changes but without explicitly finding \( k \).
02

Expressing Original Frequency

From the problem, the original mass \( m_1 \) is 0.750 kg, and the original frequency \( f_1 \) is 1.33 Hz. Substitute these into the frequency formula: \( f_1 = \frac{1}{2\pi} \sqrt{\frac{k}{m_1}} = 1.33 \).
03

Finding Frequency with Added Mass

Add 0.220 kg to the original mass: \( m_2 = 0.750 + 0.220 = 0.970 \) kg. The new frequency \( f_2 \) is given by \( f_2 = \frac{1}{2\pi} \sqrt{\frac{k}{m_2}} \). Since \( \frac{f_2}{f_1} = \sqrt{\frac{m_1}{m_2}} \), solve for \( f_2 \): \[ f_2 = f_1 \cdot \sqrt{\frac{m_1}{m_2}} = 1.33 \cdot \sqrt{\frac{0.750}{0.970}} \].
04

Calculating Added Mass Frequency

Compute \( f_2 \): \[ f_2 \approx 1.33 \times \sqrt{\frac{0.750}{0.970}} \approx 1.17 \text{ Hz} \]. This is the frequency after adding 0.220 kg to the original mass.
05

Finding Frequency by Subtracting Mass

Subtract 0.220 kg from the original mass: \( m_3 = 0.750 - 0.220 = 0.530 \) kg. Similarly, \( f_3 = \frac{1}{2\pi} \sqrt{\frac{k}{m_3}} \). Using \( \frac{f_3}{f_1} = \sqrt{\frac{m_1}{m_3}} \), solve for \( f_3 \): \[ f_3 = f_1 \cdot \sqrt{\frac{m_1}{m_3}} = 1.33 \cdot \sqrt{\frac{0.750}{0.530}} \].
06

Calculating Subtracted Mass Frequency

Compute \( f_3 \): \[ f_3 \approx 1.33 \times \sqrt{\frac{0.750}{0.530}} \approx 1.56 \text{ Hz} \]. This is the frequency after subtracting 0.220 kg from the original mass.

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Key Concepts

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

Oscillating Systems
An oscillating system is a system that moves back and forth in a regular pattern. This motion is most commonly seen in the mass-spring system, where a mass is attached to a spring and moves in a repetitive cycle. The key feature of these systems is their ability to restore themselves to an equilibrium position after being disturbed. This restorative motion is one of the fundamental properties of oscillating systems.

Oscillating systems can be seen in everyday life, from the swinging of a pendulum to the vibrations of a guitar string, each exhibiting a to-and-fro motion that repeats over time. Understanding how these systems work helps explain many natural phenomena and technological applications.
  • It involves repetitive movement over time.
  • Common examples include musical instruments, playground swings, and clocks.
  • Central to the concept is restoring forces, such as those from springs and gravity.
Exploring the dynamics of these systems involves understanding how energy is transferred and conserved as the system oscillates.
Spring Constant
The spring constant, denoted as "k", is a measure of the stiffness of a spring. It is a crucial factor in determining how a spring reacts to forces, acting as the proportionality constant in Hooke's Law. Hooke's Law states that the force exerted by a spring is proportional to its extension or compression, given by the equation:\[ F = kx \]where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.

A higher spring constant means a stiffer spring that requires more force to deform it by a certain amount. In oscillating systems like the mass-spring system, the spring constant influences the frequency of oscillation, even though this problem assumes we don't need to find the actual value of "k". Understanding the spring constant helps predict how a spring will behave when forces are applied, which is essential in designing mechanical systems.
  • The spring constant "k" determines the stiffness of a spring.
  • It is used in Hooke's Law to relate force and displacement.
  • Influences the frequency of oscillations in a mass-spring system.
Harmonic Motion
Harmonic motion is a type of precise and repetitive movement that occurs in oscillating systems, characterized by the restoring force being directly proportional to the displacement. This kind of motion is beautifully predictable and is well described by sinusoidal functions.

For a mass-spring system, the motion is termed 'simple harmonic motion', where the system oscillates with constant amplitude and period. The equation of motion for simple harmonic motion is:\[ F = -kx \]indicating that the force always acts to bring the mass back to the center or equilibrium position.

Harmonic motion forms the foundation for understanding more complex oscillations and vibrations seen in nature and engineered systems. It brings clarity to the way energy is conserved and transferred in oscillating systems. If you were to graph this motion, you'd see a wave-like pattern, making this concept essential for students studying waves and vibrations.
  • It is defined by repeated, cyclic motion.
  • Simple harmonic motion is predictable and periodic.
  • Graphs as a smooth, consistent wave pattern.
Frequency Calculation
Frequency calculation is a vital aspect of understanding oscillating systems. It refers to the number of cycles an oscillating system completes in one second. This concept is measured in Hertz (Hz). The frequency of a mass-spring system is determined by the mass on the spring and the spring constant, as shown in the formula:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]Here, \( f \) is the frequency, \( k \) is the spring constant, and \( m \) is the mass. In this context, frequency tells us how fast or slow the oscillations occur.

When you change the mass on the spring, the frequency adjusts inversely. In the exercise example, adding mass decreases the frequency, while subtracting mass increases it. This dependency is crucial for understanding how mass variations affect the energy and motion in oscillating systems.
  • Frequency measures oscillations per second, in Hertz.
  • Depends on both the spring constant and the attached mass.
  • Changing mass inversely affects the frequency.
Understanding frequency calculation is essential for designing systems that rely on precise timing and oscillation patterns, from musical instruments to precision machinery.

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Most popular questions from this chapter

A machine part is undergoing SHM with a frequency of 5.00 Hz and amplitude 1.80 \(\mathrm{cm} .\) How long does it take the part to go from \(x=0\) to \(x=-1.80 \mathrm{cm} ?\)

Inside a NASA test vehicle, a \(3.50-\mathrm{kg}\) ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m}\) . The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached 45.0 \(\mathrm{m} / \mathrm{s}\) , its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

A \(40.0 .\) N force strethes a vertical spring 0.250 \(\mathrm{m}\) . (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 \(\mathrm{s} ?(6)\) If the amplitude of the motion is 0.050 \(\mathrm{m}\) and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 \(\mathrm{s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 \(\mathrm{m}\) below the equilibrium position, moving upward?

A \(0.150-\mathrm{kg}\) toy is undergoing SHM on the end of a horizontal spring with force constant \(k=300 \mathrm{N} / \mathrm{m}\) . When the object is 0.0120 \(\mathrm{m}\) from its equilibrium position, it is observed to have a speed of 0.300 \(\mathrm{m} / \mathrm{s}\) . What are (a) the total energy of the object at any point of its motion; (b) the amplitude of the motion; (c) the maximum speed attained by the object during its motion?

The two pendulums shown in Fig 13.34 each consist of a uniform solid ball of mass \(M\) supported by a massless string, but the ball for pendulum \(A\) is very tiny while the ball for pendulum \(B\) is much larger. Find the period of each pendulum for small displacements. Which ball takes longer to complete a swing?

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