/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q109P The physical pendulum in Fig. 15... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The physical pendulum in Fig. 15-62 has two possible pivot points A and B. Point A has a fixed position but B is adjustable along the length of the pendulum as indicated by the scaling. When suspended from A, the pendulum has a period ofT=1.80s. The pendulum is then suspended from B, which is moved until the pendulum again has that period. What is the distance L between A and B?

Short Answer

Expert verified

Length between A and B is 0.804 m

Step by step solution

01

The given data

The period of oscillations, T=1.8sec.

02

Understanding the concept of SHM

The pendulum is suspended from B, but according to the problem, there is no change in the period when it is suspended from A. Sousing the basic formula of the period in terms of length and acceleration due to gravity, we can find the length between A and B.

Formula:

The period of an oscillation, T=2Ï€Lg (i)

03

Calculation of length between A and B

Though the pendulum is suspended from B, but according to the problem, there is no change in the time period. We can treat the physical pendulum as a simple pendulum with a length equal to the length from the pivot point to its center of mass.

Using equation (i) and the given value of the period, we can get the length of the pendulum between A and B is given as:

L=T2g4π2=(1.8s)2×(9.8m/s2)4π2=0.804m

Hence, the value of the length is 0.804 m.

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 91Ó°ÊÓ!

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

In Figure 15-41, block 2 of massoscillates on the end of a spring in SHM with a period of20ms.The block’s position is given byx=(1.0cm)cos(Ó¬³Ù+Ï€/2)Block 1 of mass4.0kgslides toward block 2with a velocity of magnitude6.0m/s, directed along the spring’s length. The two blocks undergo a completely inelastic collision at timet=5.0ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

When the displacement in SHM is one-half the amplitude Xm,

  1. What fraction of the total energy is kinetic energy?
  2. What fraction of the total energy is potential energy?
  3. At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

An oscillating block–spring system takes 0.75 sto begin repeating its motion.

  1. Find the period.
  2. Find the frequency in hertz.
  3. Find the angular frequency in radians per second.

In Figure 15-31, two springs are attached to a block that can oscillate over a frictionless floor. If the left spring is removed, the block oscillates at a frequency of 30 Hz. If, instead, the spring on the right is removed, the block oscillates at a frequency of 45 Hz. At what frequency does the block oscillate with both springs attached?

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?

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.