Q. 62

Expert-verified
Found in: Page 418

### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

# FIGURE P15.62 is a top view of an object of mass connected between two stretched rubber bands of length . The object rests on a frictionless surface. At equilibrium, the tension in each rubber band is . Find an expression for the frequency of oscillations perpendicular to the rubber bands. Assume the amplitude is sufficiently small that the magnitude of the tension in the rubber bands is essentially unchanged as the mass oscillates.

An expression for the frequency of oscillations perpendicular to the rubber bands is

See the step by step solution

## Step 1 Introduction

1. Newton's Second Law states that the net force acting on a mass body is proportional to the acceleration of the body

2. For a particle in simple harmonic motion, the usual equation of motion is:

Where is the motion's angular frequency.

3. The angular frequency of a particle is related to its oscillation frequency as follows:

## Step 2 Given Data

1. The mass of the object is: .

2. The length of each rubber band is: .

3. The tension in each rubber band at equilibrium is: .

## Step 3 Explanation

The figure depicts a free-body diagram for the block as it movesvertically, with denoting the rubber bands' tension force.

## Step 4 Newtons' second law

When we use Newton's second law in the vertical direction from Equation (1), we get:

## Step 5 Sine of the angle

The sine of the angle can be calculated using the geometry shown in Figure:

## Step 6 Comparing Equations

since the amplitude is supposed to be modest. As a result, we can ignore in the denominator.

Substitute for into Equation (4):

Comparing Equations (2) and (5), we obtain:

## Step 7 The frequency of oscillations

Equation (3) is then used to calculate the frequency of oscillations perpendicular to the rubber bands: