A pendulum passes \(x=0\) with a speed of \(0.50 \mathrm{m} / \mathrm{s}\); it swings out to \(A=0.20 \mathrm{m} .\) What is the period \(T\) of the pendulum? (Assume the amplitude is small.)

Short Answer

Expert verified
Answer: The period of the pendulum is approximately 2.53 seconds.

Step by step solution

01

1. Find the length of the pendulum.

First, we need to find the length of the pendulum. It is possible to get the length by utilizing the information given: When the pendulum passes through the equilibrium point, it has maximum kinetic energy, and when it is at its maximum displacement (amplitude), it has maximum potential energy. By using the conservation of energy principle, we have: \(E_{k}=E_{p}\) \(\frac{1}{2} m v^{2}=mgh\) where \(m\) is the mass of the pendulum bob, \(v\) is the initial velocity, \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^{2}}\)), and \(h\) is the height of the pendulum bob at its maximum displacement. Note that the mass of the bob (m) cancels out. Since \(A = 0.20 m\) and we assume the amplitude is small, then \(h \approx \frac{1}{2}A\). Therefore, \(\frac{1}{2} v^{2}=g(\frac{1}{2}A)\) Now, we can solve the equation to find the length \(L\) of the pendulum: \(L = \frac{h}{\sin{\theta}} \approx \frac{A}{\theta}\), \(\theta = \sin^{-1}(\frac{2h}{A})\)
02

2. Calculate the period of the pendulum.

To find the period T, we can use the formula for the period of a small-angle oscillation: \(T=2\pi\sqrt{\frac{L}{g}}\) Now, we have all the required information to calculate T.
03

3. Substitute the values and solve for T.

Substitute the values of velocity and amplitude into the equation for the length of pendulum and then use the length to find the period: \(v=0.50 \mathrm{m/s}\), \(A=0.20\mathrm{m}\) From step 1: \(\frac{1}{2} (0.50 \mathrm{m/s})^{2} = 9.8\mathrm{m/s^{2}}(\frac{1}{2}h)\) The result gives \(h = 0.01276\mathrm{m}\). Now find \(\theta\): \(\theta = \sin^{-1}(\frac{2 \times 0.01276\mathrm{m}}{0.20\mathrm{m}})\) which gives \(\theta \approx 0.1275 \, \text{rad}\). Now we can find the length \(L\): \(L \approx \frac{0.20\mathrm{m}}{0.1275} = 1.569\mathrm{m}\) Finally, calculate the period T using the formula from step 2: \(T=2\pi\sqrt{\frac{1.569\mathrm{m}}{9.8\mathrm{m/s^{2}}}}\) This gives us: \(T\approx 2.53\,\text{s}\) So the period \(T\) of the pendulum is approximately \(2.53\) seconds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Atmospheric pressure on Venus is about 90 times that on Earth. A steel sphere with a bulk modulus of 160 GPa has a volume of \(1.00 \mathrm{cm}^{3}\) on Earth. If it were put in a pressure chamber and the pressure were increased to that of Venus (9.12 MPa), how would its volume change?
The period of oscillation of a spring-and-mass system is \(0.50 \mathrm{s}\) and the amplitude is \(5.0 \mathrm{cm} .\) What is the magnitude of the acceleration at the point of maximum extension of the spring?
A steel beam is placed vertically in the basement of a building to keep the floor above from sagging. The load on the beam is $5.8 \times 10^{4} \mathrm{N},\( the length of the beam is \)2.5 \mathrm{m},$ and the cross- sectional area of the beam is \(7.5 \times 10^{-3} \mathrm{m}^{2}\) Find the vertical compression of the beam.
A baby jumper consists of a cloth seat suspended by an elastic cord from the lintel of an open doorway. The unstretched length of the cord is $1.2 \mathrm{m}\( and the cord stretches by \)0.20 \mathrm{m}$ when a baby of mass \(6.8 \mathrm{kg}\) is placed into the seat. The mother then pulls the seat down by \(8.0 \mathrm{cm}\) and releases it. (a) What is the period of the motion? (b) What is the maximum speed of the baby?
A bungee jumper leaps from a bridge and undergoes a series of oscillations. Assume \(g=9.78 \mathrm{m} / \mathrm{s}^{2} .\) (a) If a \(60.0-\mathrm{kg}\) jumper uses a bungee cord that has an unstretched length of \(33.0 \mathrm{m}\) and she jumps from a height of \(50.0 \mathrm{m}\) above a river, coming to rest just a few centimeters above the water surface on the first downward descent, what is the period of the oscillations? Assume the bungee cord follows Hooke's law. (b) The next jumper in line has a mass of \(80.0 \mathrm{kg} .\) Should he jump using the same cord? Explain.
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free