Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 55

What is the period of a pendulum consisting of a \(6.0-\mathrm{kg}\) mass oscillating on a 4.0 -m-long string?

Short Answer

Expert verified
Answer: The period of the pendulum is approximately 4.03 s.

Step by step solution

01

Identify the given values and the formula to use

We are given the mass of the pendulum (\(6.0\, \mathrm{kg}\)) and the length of the string (\(4.0\, \mathrm{m}\)). We need to find the period (T) of the pendulum. The formula to use is: \(T = 2\pi\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\).
02

Plug in the given values

Now, substitute the given values into the formula: \(T = 2\pi\sqrt{\frac{4.0\, \mathrm{m}}{9.81\, \mathrm{m/s^2}}}\).
03

Calculate the period

Calculate the square root of the ratio and then multiply by \(2\pi\): \(T \approx 2\pi\sqrt{0.408} \approx 4.03\, \mathrm{s}\).
04

Write the answer

The period of the pendulum with a \(6.0\, \mathrm{kg}\) mass oscillating on a \(4.0\, \mathrm{m}\)-long string is approximately \(4.03\, \mathrm{s}\).

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.

Start your free trial

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

The ratio of the tensile (or compressive) strength to the density of a material is a measure of how strong the material is "pound for pound." (a) Compare tendon (tensile strength \(80.0 \mathrm{MPa}\), density $1100 \mathrm{kg} / \mathrm{m}^{3}\( ) with steel (tensile strength \)\left.0.50 \mathrm{GPa}, \text { density } 7700 \mathrm{kg} / \mathrm{m}^{3}\right)$ which is stronger "pound for pound" under tension? (b) Compare bone (compressive strength \(160 \mathrm{MPa}\), density $1600 \mathrm{kg} / \mathrm{m}^{3}$ ) with concrete (compressive strength $\left.0.40 \mathrm{GPa}, \text { density } 2700 \mathrm{kg} / \mathrm{m}^{3}\right):$ which is stronger "pound for pound" under compression?
A brass wire with Young's modulus of \(9.2 \times 10^{10} \mathrm{Pa}\) is $2.0 \mathrm{m}\( long and has a cross-sectional area of \)5.0 \mathrm{mm}^{2} .$ If a weight of \(5.0 \mathrm{kN}\) is hung from the wire, by how much does it stretch?
A mass-spring system oscillates so that the position of the mass is described by \(x=-10 \cos (1.57 t),\) where \(x\) is in \(\mathrm{cm}\) when \(t\) is in seconds. Make a plot that has a dot for the position of the mass at $t=0, t=0.2 \mathrm{s}, t=0.4 \mathrm{s}, \ldots, t=4 \mathrm{s}$ The time interval between each dot should be 0.2 s. From your plot, tell where the mass is moving fastest and where slowest. How do you know?
Equipment to be used in airplanes or spacecraft is often subjected to a shake test to be sure it can withstand the vibrations that may be encountered during flight. A radio receiver of mass \(5.24 \mathrm{kg}\) is set on a platform that vibrates in SHM at \(120 \mathrm{Hz}\) and with a maximum acceleration of $98 \mathrm{m} / \mathrm{s}^{2}(=10 \mathrm{g}) .$ Find the radio's (a) maximum displacement, (b) maximum speed, and (c) the maximum net force exerted on it.
The maximum strain of a steel wire with Young's modulus $2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2},\( just before breaking, is \)0.20 \%$ What is the stress at its breaking point, assuming that strain is proportional to stress up to the breaking point?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Thermodynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Engineering Physics

Read Explanation

Linear Momentum

Read Explanation
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