Question

FIGURE P15.62 is a top view of an object of mass $m$ connected between two stretched rubber bands of length $L$. The object rests on a frictionless surface. At equilibrium, the tension in each rubber band is $T$. 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.

Short answer

An expression for the frequency of oscillations perpendicular to the rubber bands is $f=\frac{1}{2\pi }\sqrt{\frac{2T}{mL}}$

Step-by-step solution

Introduction

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

$\sum F=ma$

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

$\frac{d^{2}y}{d^2}=-{\omega }^{2}y$

Where $\omega$is the motion's angular frequency.

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

$f=\frac{\omega }{2\pi }$

Given Data

1. The mass of the object is: $m$.

2. The length of each rubber band is: $L$.

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

Explanation

The figure depicts a free-body diagram for the block as it moves$y$vertically, with $T$denoting the rubber bands' tension force.

Newtons' second law 

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

$\sum F_y=-2T\sin \theta =m{a}_y$

$-2T\sin \theta =m\frac{d^{2}y}{d{t}^2}$

Sine of the angle

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

localid="1650092164005" $\sin \theta =\frac{y}{\sqrt{y^2+L^2}}$

Comparing Equations

$y⩽L$since the amplitude is supposed to be modest. As a result, we can ignore $y^2$in the denominator.

$\sin \theta =\frac{y}{\sqrt{L^2}}=\frac{y}{L}$

Substitute for $\sin \theta$into Equation (4):

$m\frac{d^{2}y}{d{t}^2}=-2T\frac{y}{L}$

$\frac{d^{2}y}{d{t}^2}=-\left(\frac{2T}{mL}\right)y$

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

${\omega }^2=\frac{2T}{mL}$

$\omega =\sqrt{\frac{2T}{mL}}$

The frequency of oscillations 

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

$f=\frac{\omega }{2\pi }$

$=\frac{1}{2\pi }\sqrt{\frac{2T}{mL}}$