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.

Short Answer

Expert verified
Answer: The period of oscillation without damping is approximately 2.01 seconds, and the time for the amplitude to reduce with damping is approximately 6.93 seconds.

Step by step solution

01

Determine the Period of Oscillation Without Damping

To find the period of the oscillation without any damping, we'll use the formula for the period of a simple pendulum: \(T = 2\pi\sqrt{\frac{L}{g}}\) where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{m/s^2}\)). Plugging in the values, we get: \(T = 2\pi\sqrt{\frac{1.00 \mathrm{m}}{9.81 \mathrm{m/s^2}}}\)
02

Calculate the Period of Oscillation Without Damping

To find the period of oscillation without damping, we can now solve the equation obtained in step 1: \(T \approx 2\pi\sqrt{\frac{1.00 \mathrm{m}}{9.81 \mathrm{m/s^2}}} \approx 2\pi\sqrt{0.102} \approx 2.01 \mathrm{s}\) This is the period of oscillation without any damping.
03

Determine the Time for the Amplitude to Reduce With Damping

Now we have to find the time for the amplitude to reduce in the presence of a damping constant \(b = 0.0100 \mathrm{kg/s}\). The amplitude with damping is given by: \(A(t) = A_0 e^{-\frac{b}{2m}t}\) We are given the initial amplitude \(A_0 = 10.0^{\circ}\) and want to find the time \(t\) when the amplitude reduces to \(5.0^{\circ}\). Let's first convert the angles to radians: \(A_0 = \frac{10}{180}\pi \mathrm{rad}\) \(A(t) = \frac{5}{180}\pi \mathrm{rad}\) Now we can write the equation using these amplitudes: \(\frac{5}{180}\pi \mathrm{rad} = \frac{10}{180}\pi \mathrm{rad} e^{-\frac{0.0100}{2(0.050)}t}\)
04

Solve for the Time for the Amplitude to Reduce With Damping

We can now solve the equation in step 3 for \(t\): \(\frac{\frac{5}{180}\pi \mathrm{rad}}{\frac{10}{180}\pi \mathrm{rad}} = e^{-\frac{0.0100}{2(0.050)}t}\) \(\frac{1}{2} = e^{-\frac{1}{10}t}\) Take the natural logarithm of both sides: \(-\ln{2} = -\frac{1}{10}t\) And finally, solve for \(t\): \(t = \frac{-10\ln{2}}{-1} \approx 6.93 \mathrm{s}\) #Conclusion# The period of oscillation without damping is approximately \(2.01 \mathrm{s}\), and the time for the amplitude to reduce to \(5.0^{\circ}\) with damping is approximately \(6.93 \mathrm{s}\).

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Key Concepts

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

Simple Pendulum Period
In physics, the simple pendulum period refers to the time it takes for a pendulum to complete one full swing from one side to the other and back again. This period is represented by the symbol T and is crucial for understanding the harmonic motion of pendulums. For a pendulum swinging with small angles (less than approximately 15 degrees), the period is independent of the amplitude of the swing.

The formula for the period of a simple pendulum is given by:
\[T = 2\pi\sqrt{\frac{L}{g}}\]
where L is the length of the pendulum and g is the acceleration due to gravity. This relationship suggests that the period is only dependent on the length of the pendulum and the gravitational field strength, but not on the mass of the bob or the amplitude of the swing. This is why in the textbook solution, once the length L is known, the calculation of the period becomes straightforward.
Damped Oscillation
When a pendulum or any other oscillator moves in a medium that offers resistance, such as air or liquid, its movement is not perfectly harmonic. Instead, it experiences a damped oscillation. Damping is caused by non-conservative forces like friction and air resistance, which remove energy from the system, causing the amplitude of the oscillation to decrease over time. For small damping, which is often the case in introductory physics problems, the effect on the period is minimal.

The degree of damping is characterized by the damping constant b, with higher values of b indicating stronger damping effects. In our textbook exercise, by including a damping constant, we can compute various dynamics of the damped pendulum, such as the time it would take for the amplitude of oscillation to reduce to half its initial value.
Exponential Decay in Amplitude
The reduction in the amplitude of a damped oscillation follows the pattern of exponential decay. This means that the amplitude decreases at a rate proportional to its current value, leading to a rapid drop initially, which gradually slows down as the amplitude becomes smaller.

The mathematical expression for the amplitude of a damped oscillator as a function of time is:
\[A(t) = A_0 e^{-\frac{b}{2m}t}\]
where A_0 is the initial amplitude, m is the mass of the oscillator, b is the damping constant, and t is the time. The factor \(-\frac{b}{2m}\) in the exponent is the damping ratio, and it determines the rate of exponential decay. Exponential decay is also used in other physics contexts such as radioactive decay and RC circuits.

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Most popular questions from this chapter

A shock absorber that provides critical damping with \(\omega_{\gamma}=72.4 \mathrm{~Hz}\) is compressed by \(6.41 \mathrm{~cm} .\) How far from the equilibrium position is it after \(0.0247 \mathrm{~s} ?\)

Two pendulums of identical length of \(1.000 \mathrm{~m}\) are suspended from the ceiling and begin swinging at the same time. One is at Manila, in the Philippines, where \(g=9.784 \mathrm{~m} / \mathrm{s}^{2}\) and the other is at Oslo, Norway, where \(g=9.819 \mathrm{~m} / \mathrm{s}^{2},\) After how many oscillations of the Manila pendulum will the two pendulums be in phase again? How long will it take for them to be in phase again?

Two children are on adjacent playground swings of the same height. They are given pushes by an adult and then left to swing. Assuming that each child on a swing can be treated as a simple pendulum and that friction is negligible, which child takes the longer time for one complete swing (has a longer period)? a) the bigger child d) the child given the b) the lighter child biggest push c) neither child

14.10 A pendulum is suspended from the ceiling of an elevator. When the elevator is at rest, the period of the pendulum is \(T\). The elevator accelerates upward, and the period of the pendulum is then a) still T. b) less than \(T_{-}\) c) greater than \(T\).

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.

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