   Chapter 13, Problem 30P

Chapter
Section
Textbook Problem

An object executes simple harmonic motion with an amplitude A. (a) At what values of its position does its speed equal half its maximum speed? (b) At what values of its position does its potential energy equal half the total energy?

a)

To determine
The position of the object executing simple harmonic oscillator where the velocity is half of the maximum velocity.

Explanation

At any instant of time, the total energy of the simple harmonic oscillator is,

E=12mv2+12kx2

Here,

E is the total energy

m is the mass of the object

v is the velocity of the object

k is the force constant of the spring

x is the displacement from equilibrium position

When the system is in equilibrium position, the total energy is in the form of kinetic energy.

E=12mvmax2

Here,

vmax is the maximum velocity of the object

When the system is in the turning points, the total energy is in the form of elastic potential energy.

E=12kA2

Here,

A is the amplitude of the motion

From the above equations,

kA2=mvmax2

On re-arranging

vmax2=kA2m

When

b)

To determine
The position where its potential energy is equal half the total energy.

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