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Chapter 15: Q115P (page 443)

What is the length of a simple pendulum whose full swing from left to right and then back again takes 3.2 S?

Short Answer

Expert verified

The length of a simple pendulum is 2.54.

Step by step solution

01

The given data

Full swing time is3.2 s .

02

Understanding the concept of the period of Simple Harmonic Motion

Using the formula for the period of a simple pendulum, we can find the length of a simple pendulum.

Formula:

The time period of a simple pendulum in SHM,

T=2πLg

03

Calculation of the length of the pendulum

Using equation (i), we have the value of the length of the pendulum as:
L=T2g4π=3.2s2×9.8m/s24π2=2.54m

Hence, the required value of length is 2.54m.

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Most popular questions from this chapter

Figure below gives the position of a 20 gblock oscillating in SHM on the end of a spring. The horizontal axis scale is set byts=40.0ms.

  1. What is the maximum kinetic energy of the block?
  2. What is the number of times per second that maximum is reached? (Hint: Measuring a slope will probably not be very accurate. Find another approach.)

What is the frequency of a simple pendulum 2.0mlong (a) in a room, (b) in an elevator accelerating upward at a rate of role="math" localid="1657259780987" 2.0m/s2, and (c) in free fall?

When a 20 Ncan is hung from the bottom of a vertical spring, it causes the spring to stretch 20 cm .

  1. What is the spring constant?
  2. This spring is now placed horizontally on a frictionless table. One end of it is held fixed, and the other end is attached to a 5.0 Ncan. The can is then moved (stretching the spring) and released from rest. What is the period of the resulting oscillation?

A block of massM=5.4kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constantk=6000N/m. A bullet of massm=9.5gand velocityv→of magnitud630m/sstrikes and is embedded in the block (SeeFigure). Assuming the compression of the spring is negligible until the bullet is embedded.

(a) Determine the speed of the block immediately after the collision and

(b) Determine the amplitude of the resulting simple harmonic motion.

For Equationx=xmcos(ωt+ϕ), suppose the amplitudexmis given by

xm=Fm[m2(ωd2−ω2)2+b2ωd2]1/2

whereFmis the (constant) amplitude of the external oscillating force exerted on the spring by the rigid support in Figure below. At resonance,

  1. what is the amplitude of the oscillating object?
  2. what is the velocity amplitude of the oscillating object?
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