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Chapter 13: Problem 49

Derive the period of a simple pendulum by considering the horizontal displacement \(x\) and the force acting on the bob, rather than the angular displacement and torque.

Short Answer

Expert verified
The period \(T\) of a simple pendulum, considering the horizontal displacement and the acting force on the bob, derived from the equation of motion, is \(T = 2\pi \sqrt{{L}/{g}}\).

Step by step solution

01

Identify the forces acting on the pendulum

The forces acting on a pendulum bob are gravity and tension. Decompose these vectors into the horizontal and vertical components. The vertical force component is \(mg \cos \theta\) and the horizontal force component is \(mg \sin \theta \). Because the pendulum is at rest in the vertical direction the vertical component\(T \cos \theta\) of the tension cancels exactly with the weight, so \(T = mg / \cos \theta\). Hence, we are mainly interested in the horizontal force which creates a restoring force.
02

Relation between horizontal displacement, length and angle

For small angles, we can approximate \(\sin \theta \approx \theta\), where \(\theta\) is actually \(\theta = x / L\), with \(x\) being the horizontal displacement and \(L\) the length of the pendulum. Thus we have \(mg \sin \theta = mg \theta = mg x/ L\).
03

Determine the equation of motion

Applying Newton's second law in the horizontal direction, the net force in that direction is equal to the mass times acceleration in that direction. So we can equate \(mgx / L\) to \(ma\), and because \(a\) is actually the second derivative of the displacement with respect to time, we get \(mgx / L = m (d^2 x / dt^2)\).
04

Solve the differential equation

Now we have a second-order linear homogeneous differential equation: \((d^2 x / dt^2) + g/L \cdot x = 0\). The solution to this equation is in the form \(x(t) = A cos(\sqrt{g/L} \cdot t + \phi)\), where \(A\) and \(\phi\) are constants that depend on initial conditions. The form \(x(t) = A cos(\sqrt{g/L} \cdot t + \phi)\) is the general solution to a simple harmonic oscillator with frequency \(f = 1/(2 \pi) \sqrt{g/L}\). Hence the period \(T\) of the pendulum is the inverse of the frequency.

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Most popular questions from this chapter

A physics student, bored by a lecture on simple harmonic moion, idly picks up his pencil (mass \(8.65 \mathrm{~g}\), length \(18.8 \mathrm{~cm}\) ) by the tip with his frictionless fingers, and allows it to swing back and forth with small amplitude. If the pencil completes 5974 full cycles during the lecture, how long does the lecture last?

A pendulum of length \(L\) is mounted in a rocket. Find expressions for its period if the rocket is (a) at rest on its launch pad, (b) accelerating upward with acceleration \(a=\frac{1}{2} g\), (c) accelerating downward with \(a=\frac{1}{2} g\), and (d) in free fall.

Show that \(x(t)=a \cos \omega t-b \sin \omega t\) represents simple harmonic motion, as in Equation \(13.8\), with \(A=\sqrt{a^{2}+b^{2}}\) and \(\phi=\tan ^{-1}(b / a)\).

The protein dynein powers the flagella that propel some unicellular organisms. Biophysicists have found that dynein is intrinsically oscillatory, and that it exerts peak forces of about \(1.0 \mathrm{pN}\) when it attaches to structures called microtubules. The resulting oscillations have amplitude \(16 \mathrm{~nm}\). (a) If this system is modeled as a mass-spring system, what's the associated spring constant? (b) If the oscillation frequency is \(72 \mathrm{~Hz}\), what's the effective mass?

an ornithologist who knows you've studied physics. She asks you for a noninvasive way to measure birds" masses. You propose using a bird feeder in the shape of a 50 -cm-diameter disk of mass \(340 \mathrm{~g}\), suspended by a wire with torsional constant \(5.00 \mathrm{~N} \cdot \mathrm{m} / \mathrm{rad}(\mathrm{Fig} .13 .35)\). Two birds land on opposite sides and the feeder goes into torsional oscillation at \(2.6 \mathrm{~Hz}\). Assuming the birds have the same mass, what is it?
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