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An object of mass \(m\) is hung from the base of an ideal spring that is suspended from the ceiling. The spring has a spring constant \(k .\) The object is pulled down a distance \(D\) from equilibrium and released. Later, the same system is set oscillating by pulling the object down a distance \(2 D\) from equilibrium and then releasing it. (a) How do the period and frequency of oscillation change when the initial displacement is increased from \(D\) to $2 D ?$ (b) How does the total energy of oscillation change when the initial displacement is increased from \(D\) to \(2 D ?\) Give the answer as a numerical ratio. (c) The mass-spring system is set into oscillation a third time. This time the object is pulled down a distance of \(2 D\) and then given a push downward some more, so that it has an initial speed \(v_{i}\) downward. How do the period and frequency of oscillation compare to those you found in part (a)? (d) How does the total energy compare to when the object was released from rest at a displacement \(2 D ?\)

Step by step solution

01

(a) Comparing periods and frequencies of oscillation

Since the mass-spring system is ideal, the period of oscillation does not depend on the amplitude of oscillation (displacement). The period formula for a mass-spring system is given by: $$T=2\pi\sqrt{\frac{m}{k}}$$ Since neither the mass nor the spring constant changes between the two displacements, the period (T) remains the same for both cases. Since frequency (f) is the inverse of the period: $$f=\frac{1}{T}$$ The frequency also remains the same for both cases. ——————————————————

02

(b) Ratio of total energy of oscillation

The total energy of oscillation for a mass-spring system is given by the potential energy stored in the spring: $$E=\frac{1}{2}kx^2$$ We need to find the ratio of total energy when the initial displacement is increased from D to 2D. Let \(E_1\) be the energy for displacement D and \(E_2\) for displacement 2D: $$E_1=\frac{1}{2}kD^2$$ $$E_2=\frac{1}{2}k(2D)^2=\frac{1}{2}k(4D^2)$$ The ratio of the total energies is: $$\frac{E_2}{E_1}=\frac{\frac{1}{2}k(4D^2)}{\frac{1}{2}kD^2}=4$$ Thus, the total energy of oscillation increases by a factor of 4 when the initial displacement is doubled from D to 2D. ——————————————————

03

(c) Comparing period and frequency of oscillation with initial velocity

As before, the period and frequency of the mass-spring system do not depend on the amplitude of oscillation or the initial velocity. Therefore, even with an initial velocity downward, the period and frequency of oscillation will be the same as those found in part (a). ——————————————————

04

(d) Comparing total energy with initial velocity

When the object is released from rest at a displacement of 2D with an initial speed \(v_{i}\) downward, the total mechanical energy of the system will have an additional contribution from the kinetic energy: $$E_{total}=E_{potential}+E_{kinetic}$$ For the case of initial speed \(v_{i}\) downward: $$E_{total}=\frac{1}{2}k(2D)^2+\frac{1}{2}mv_{i}^2$$ For the case where the object was released from rest at a displacement of 2D: $$E_{total}=\frac{1}{2}k(2D)^2$$ We cannot find the exact ratio of energies without knowing the initial velocity \(v_{i}\). However, we can say that the total energy will be higher when the object is given an initial downward velocity, as it has an additional contribution from kinetic energy.

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