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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

An \(80.0-\mathrm{kg}\) bungee jumper is enjoying an afternoon of jumps. The jumper's first oscillation has an amplitude of \(10.0 \mathrm{~m}\) and a period of \(5.00 \mathrm{~s}\). Treating the bungee cord as spring with no damping, calculate each of the following: a) the spring constant of the bungee cord. b) the bungee jumper's maximum speed during the ascillation, and c) the time for the amplitude to decrease to \(2.00 \mathrm{~m}\) (with air resistance providing the damping of the oscillations at \(7.50 \mathrm{~kg} / \mathrm{s})\)

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The motion of a planet in a circular orbit about a star obeys the equations of simple harmonic motion. If the orbit is observed edge-on, so the planet's motion appears to be onedimensional, the analogy is quite direct: The motion of the planet looks just like the motion of an object on a spring a) Use Kepler's Third Law of planetary motion to determine the "spring constant" for a planet in circular orbit around a star with period \(T\) b) When the planet is at the extremes of its motion observed edge-on, the analogous "spring" is extended to its largest displacement. Using the "spring" analogy, determine the orbital velocity of the planet.

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

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