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Chapter 14: Problem 6

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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Most popular questions from this chapter

A block of wood of mass \(55.0 \mathrm{~g}\) floats in a swimming pool, oscillating up and down in simple harmonic motion with a frequency of \(3.00 \mathrm{~Hz}\). a) What is the value of the effective spring constant of the water? b) A partially filled water bottle of almost the same size and shape as the block of wood but with mass \(250 . g\) is placed on the water's surface. At what frequency will the bottle bob up and down?

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

Consider two identical oscillators, each with spring constant \(k\) and mass \(m\), in simple harmonic motion. One oscillator is started with initial conditions \(x_{0}\) and \(v_{j}\) the other starts with slightly different conditions, \(x_{0}+\delta x\) and \(v_{0}+\delta v_{1}\) a) Find the difference in the oscillators' positions, \(x_{1}(t)-x_{2}(t)\) for all t. b) This difference is bounded; that is, there exists a constant \(C\) independent of time, for which \(\left|x_{1}(t)-x_{2}(t)\right| \leq C\) holds for all \(t\). Find an expression for \(C\). What is the best bound, that is, the smallest value of \(C\) that works? (Note: An important characteristic of chaotic systems is exponential sensitivity to initial conditions; the difference in position of two such systems with slightly different initial conditions grows exponentially with time. You have just shown that an oscillator in simple harmonic motion is not a chaotic system.)

With the right choice of parameters, a damped and driven physical pendulum can show chaotic motion, which is sensitively dependent on the initial conditions. Which statement about such a pendulum is true? a) Its long-term behavior can be predicted. b) Its long-term behavior is not predictable. c) Its long-term behavior is like that of a simple pendulum of equivalent length. d) Its long-term behavior is like that of a conical pendulum. e) None of the above is true.

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?
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