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A spring is hanging from the ceiling with a mass attached to it. The mass is pulled downward, causing it to oscillate vertically with simple harmonic motion. Which of the following will increase the frequency of oscillation? a) adding a second, identical spring with one end attached to the mass and the other to the ceiling b) adding a second, identical spring with one end attached to the mass and the other to the floor c) increasing the mass d) adding both springs, as described in (a) and (b)

Short Answer

Expert verified
A) Adding a second, identical spring with one end attached to the mass and the other to the ceiling B) Adding a second, identical spring with one end attached to the mass and the other to the floor C) Increasing the mass D) Adding both springs, as described in (A) and (B) Answer: A) Adding a second, identical spring with one end attached to the mass and the other to the ceiling

Step by step solution

01

Option A: Adding a second, identical spring with one end attached to the mass and the other to the ceiling

When a second, identical spring is added to the mass and attached to the ceiling in parallel, the effective spring constant of the system becomes \(k_\mathrm{effective} = k_1 + k_2\), where \(k_1\) and \(k_2\) are the spring constants of the two springs. Since the springs are identical, \(k_\mathrm{effective} = 2k\). The frequency will now depend on the effective spring constant and can be represented as \(f_A = \frac{1}{2\pi}\sqrt{\frac{2k}{m}}\). Comparing this to the initial frequency, we can see that the frequency has increased.
02

Option B: Adding a second, identical spring with one end attached to the mass and the other to the floor

When a second, identical spring is attached to the mass and the floor, the second spring will be in series with the first one . The effective spring constant of the system can be determined using the formula: \(\frac{1}{k_\mathrm{effective}} = \frac{1}{k_1} + \frac{1}{k_2}\). As the springs are identical, we have \(k_\mathrm{effective} = \frac{k}{2}\). The frequency is now given by \(f_B = \frac{1}{2\pi}\sqrt{\frac{k/2}{m}}\). Comparing it with the initial frequency, we see that the frequency will decrease in this case.
03

Option C: Increasing the mass

As we can see from the frequency formula \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\), increasing the mass will result in a decrease in the frequency of oscillation.
04

Option D: Adding both springs, as described in (A) and (B)

In this situation, we have a parallel setup with the ceiling (Option A) and a series setup with the floor (Option B). The result of Option A was to increase the frequency, while the result of Option B was to decrease the frequency. When we add both springs, their effects will counteract each other, and the overall effect on the frequency of oscillation will not be an increase. In conclusion, only Option A (adding a second, identical spring with one end attached to the mass and the other to the ceiling) will increase the frequency of oscillation.

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Key Concepts

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

Spring Constant
In simple harmonic motion, the spring constant, often denoted as \(k\), plays a crucial role in determining how a spring system behaves. Think of the spring constant as a measure of stiffness. A higher spring constant indicates a stiffer spring, meaning it requires more force to stretch or compress it. This concept is essential in understanding how springs affect oscillation frequency. When two identical springs are connected in parallel, their spring constants add together. For example:
  • For two springs in parallel: \(k_\text{effective} = k_1 + k_2\)
  • If both springs have the same spring constant \(k\), then \(k_\text{effective} = 2k\)
Increasing \(k_\text{effective}\) leads to a higher oscillation frequency, meaning the system will oscillate faster. This is because the system exerts more force against displacement, causing quicker movements back to equilibrium.
Understanding the connection between spring constants and oscillation helps in analyzing real-world systems such as vehicle suspensions or even heartbeats, where elasticity and regular cycles are important.
Oscillation Frequency
Oscillation frequency refers to how often an oscillating system completes a full cycle of motion in one second. In the context of springs in simple harmonic motion, frequency can be determined using:
  • Frequency formula: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\)

Where \(k\) is the spring constant and \(m\) is the mass attached to the spring. This formula reveals a few critical points:
  • A stronger spring (higher \(k\)) results in a higher frequency.
  • A larger mass (higher \(m\)) leads to a lower frequency.
Option A from the exercise, where a second spring is added in parallel, demonstrates how increasing the effective spring constant boosts the frequency. The increased stiffness accelerates the system’s return to its equilibrium position. Conversely, more mass means the system is harder to move, thus slowing oscillation and reducing frequency.
This principle is vital in engineering and science, where controlling motion frequency can lead to systems that are more efficient, more stable, or produce desired effects in manufacturing processes.
Parallel and Series Springs
Understanding how springs behave in parallel and series combinations extends their utility in various mechanical systems.

Parallel Springs:
  • Much like adding resistors in parallel, adding springs in parallel increases the system's overall stiffness.
  • The formula \(k_{\text{parallel}} = k_1 + k_2\) applies, leading to an increased effective spring constant.
  • This setup causes the system to oscillate at a higher frequency if all other factors remain constant.


Series Springs:
  • When springs are linked in series, the system as a whole becomes more flexible.
  • The reciprocal of the effective spring constant is the sum of the reciprocals of each spring: \(\frac{1}{k_{\text{series}}} = \frac{1}{k_1} + \frac{1}{k_2}\).
  • This setup generally reduces the system’s effective spring constant and lowers the oscillation frequency.
In the exercise, in scenario B, when a second spring is added in series, it demonstrates the decrease in effective stiffness, hence lowering the frequency of oscillation.
Understanding these principles is fundamental in designing devices that require precise control over motion, such as in clock mechanisms or vibration isolators.

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Most popular questions from this chapter

A mass \(M=0.460 \mathrm{~kg}\) moves with an initial speed \(v=3.20 \mathrm{~m} / \mathrm{s}\) on a level frictionless air track. The mass is initially a distance \(D=0.250 \mathrm{~m}\) away from a spring with \(k=\) \(840 \mathrm{~N} / \mathrm{m}\), which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance \(d\), before reversing direction. After bouncing off the spring the mass travels with the same speed \(v\), but in the opposite dircction. a) Determine the maximum distance that the spring is compressed. b) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)

An object in simple harmonic motion is isochronous, meaning that the period of its oscillations is independent of their amplitude. (Contrary to a common assertion, the operation of a pendulum clock is not based on this principle. A pendulum clock operates at fixed, finite amplitude. The gearing of the clock compensates for the anharmonicity of the pendulum.) Consider an oscillator of mass \(m\) in one-dimensional motion, with a restoring force \(F(x)=-c x^{3}\) where \(x\) is the displacement from equilibrium and \(c\) is a constant with appropriate units. The motion of this ascillator is periodic but not isochronous. a) Write an expression for the period of the undamped oscillations of this oscillator. If your expression involves an integral, it should be a definite integral. You do not need to evaluate the expression. b) Using the expression of part (a), determine the dependence of the period of oscillation on the amplitude. c) Generalize the results of parts (a) and (b) to an oscillator of mass \(m\) in one-dimensional motion with a restoring force corresponding to the potential energy \(U(x)=\gamma|x| / \alpha\), where \(\alpha\) is any positive value and \(\gamma\) is a constant

A mass of \(10.0 \mathrm{~kg}\) is hanging by a steel wire \(1.00 \mathrm{~m}\) long and \(1.00 \mathrm{~mm}\) in diameter. If the mass is pulled down slightly and released, what will be the frequency of the resulting oscillations? Young's modulus for steel is \(2.0 \cdot 10^{11} \mathrm{~N} / \mathrm{m}^{2}\)

A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass \(M\) and length \(L\) that is pivoted freely about one end. with a solid sphere of the same mass, \(M,\) and a radius of \(L / 2\) centered about the free end of the rod. a) Obtain an expression for the moment of inertia of the pendulum about its pivot point as a function of \(M\) and \(L\). b) Obtain an expression for the period of the pendulum for small oscillations.

A small mass, \(m=50.0 \mathrm{~g}\), is attached to the end of a massless rod that is hanging from the ceiling and is free to swing. The rod has length \(L=1.00 \mathrm{~m} .\) The rod is displaced \(10.0^{\circ}\) from the vertical and released at time \(t=0\). Neglect air resistance. What is the period of the rod's oscillation? Now suppose the entire system is immersed in a fluid with a small damping constant, \(b=0.0100 \mathrm{~kg} / \mathrm{s},\) and the rod is again released from an initial displacement angle of \(10.0^{\circ}\). What is the time for the amplitude of the oscillation to reduce to \(5.00^{\circ}\) ? Assume that the damping is small Also note that since the amplitude of the oscillation is small and all the mass of the pendulum is at the end of the rod, the motion of the mass can be treated as strictly linear, and you can use the substitution \(R \theta(t)=x(t),\) where \(R=1.0 \mathrm{~m}\) is the length of the pendulum rod.

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