A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal...


A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at t = 0 from the position x = 0.0300 m.

(a) Determine the location of the mass at t = 5.51 s?

(b) Determine if the mass is moving in the positive or negative x-direction at t = 5.51 s?

Angular Frequency:

The angular frequency is calculated by taking the ratio of the time period given for the rotation to twice the constant value of pie and expressed in radians per second. The Angular frequency is a scalar quantity which is a measure of the rate of rotation.

Answer and Explanation:

Given data

  • The time period is given as: {eq}T = 2.25\,{\rm{s}}{/eq}
  • The initial position is given as: {eq}{x_o} = 0.0300\,{\rm{m}}{/eq}
  • The time given is : {eq}t = 5.51\,{\rm{m}}{/eq}


The expression to calculate the position of the mass at given time {eq}t = 5.51\,{\rm{s}}{/eq} ,

{eq}x\left( t \right) = {x_o}\cos \left( {\omega t} \right)\,.....\,(I){/eq}

Here, {eq}\omega {/eq} is the angular frequency.

Calculate the Angular frequency.

{eq}\begin{align*} \omega & = \dfrac{{2\pi }}{T}\\ \omega & = \dfrac{{2 \times 3.14}}{{2.25}}\\ \omega & = 2.8\,{\rm{rad/s}} \end{align*}{/eq}

Substituting the value in the expression (I),

{eq}\begin{align*} x\left( {5.51} \right)& = 0.0300cos\left( {2.8 \times 5.51 \times \dfrac{{180}}{\pi }} \right)\\ x\left( {5.51} \right) &= 0.0300 \times \left( { - 0.9612} \right)\\ x\left( {5.51} \right) &= - 0.0288\,{\rm{m}} \end{align*}{/eq}

Thus, the position is {eq}0.0288\,{\rm{m}}{/eq}


The negative sign in the position of mass depicting that the mass is moving in the negative x-direction at {eq}t = 5.51\,{\rm{s}}{/eq}.

Learn more about this topic:

Linear Velocity: Definition & Formula

from MCAT Prep: Help and Review

Chapter 14 / Lesson 12

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