Question

A$100g$block attached to a spring with spring constant $2.5n\backslash m$oscillates horizontally on a frictionless table. Its velocity is $20\backslash cm$when $x=-5.0cm$

$a.$What is the amplitude of oscillation?

$b.$What is the block's maximum acceleration?

$c.$What is the block's position when the acceleration is maximum?

$d.$What is the speed of the block when$x=30cm?$

Short answer

$a.$The amplitude of oscillation is $A=0.064\mathrm{m}$

$b.$The block maximum acceleration is $a_{\max }=1.60\mathrm{m}/{\mathrm{s}}^2$

$c.$$x=-0.064\mathrm{m}$is the position of the acceleration maximum'

$d.$The speed of the block when $x=30cm$is$v=0.283\mathrm{m}/\mathrm{s}$

Step-by-step solution

Concept and principle.

- The total energy of a simple harmonic oscillator is a constant of the motion and is given by

$E=\frac{1}{2}k{A}^2$ $\left(1\right)$

- The kinetic and potential energies for an object of mass $m$oscillating at the end of a spring of force constant $k$are given by

$K=\frac{1}{2}m{v}^2$ $\left(2\right)$

$U=\frac{1}{2}k{x}^2$ $\left(3\right)$

- The angular frequency of an oscillator in simple harmonic motion is given by

$\omega =\sqrt{\frac{k}{m}}$ $\left(4\right)$

Note that $\omega$does not depend on the amplitude but only on the mass $m$and the force constant $k$

- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed $\omega$and the amplitude $a$as follows:

$a_{\max }={\omega }^{2}A$ $\left(5\right)$

Step2; Given data.

• The mass of the block is$m=(100\mathrm{g})\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=0.1\mathrm{kg}$
• The spring constant of the spring is$k=2.5\mathrm{N}/\mathrm{m}$
• The velocity of the block at $x=-(5.0\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=-0.05\mathrm{m}$is$v=(20\mathrm{cm}/\mathrm{s})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.2\mathrm{m}/\mathrm{s}$
• In part ($a$), we are asked to determine the amplitude of oscillation of the block.
• In part ($b$), we are asked to determine the block's maximum acceleration.
• In part ($c$), we are asked to determine the position of the block for which the acceleration is a maximum.
• In part ($d$), we are asked to determine the speed of the block at$x=3.0cm$

Step3; Solution of part a

Since the energy of the oscillator is conserved, the total energy of the system is constant and is equal to the sum of the kinetic and potential energies of the oscillator:

$E=K+U$

Substitute for $E$from Equation $\left(1\right)$, for $K$from Equation ($2$), and for $U$from Equation ($3$):

$\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$

$k{A}^2=m{v}^2+k{x}^2$ $\left(6\right)$

Solve for $A$:

$A=\sqrt{\frac{m{v}^2+k{x}^2}{k}}$

Substitute numerical values:

$A=\sqrt{\frac{(0.1\mathrm{kg})(0.2\mathrm{m}/\mathrm{s}{)}^2+(2.5\mathrm{N}/\mathrm{m}\left)\right(-0.05\mathrm{m}{)}^2}{2.5\mathrm{N}/\mathrm{m}}}$

$=0.064m$

Solution for part b

The angular frequency of the oscillator is found from Equation ($4$):

$\omega =\sqrt{\frac{k}{m}}$

Substitute numerical values:

$\omega =\sqrt{\frac{2.5\mathrm{N}/\mathrm{m}}{0.1\mathrm{kg}}}$

$=5^{s-1}$

Solution for equation

The maximum acceleration of the block is found from Equation ($5$):

$a_{\max }={\omega }^{2}A$

Substitute numerical values:

$a_{\max }={\left(5{\mathrm{s}}^{-1}\right)}^2(0.064\mathrm{m})$

$=1.60\mathrm{m}/{\mathrm{s}}^2$

Solution for c

The acceleration of the block has a maximum magnitude and a positive value when the block is at its most negative displacement from the equilibrium position; at negative the amplitude. So

$x=-A=-0.064\mathrm{m}$

Solution for d

Rearrange Equation ($6$) and solve it for$v$:

$m{v}^2=k{A}^2-k{x}^2$

$\mathit{m}{\mathit{v}}^{\mathbf{2}}\mathbf{=}\mathit{k}\left(A^2-x^2\right)$

$v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}$

Substitute numerical values:

$v=\sqrt{\frac{(2.5\mathrm{N}/\mathrm{m})\left[(0.064\mathrm{m}{)}^2-(0.03\mathrm{m}{)}^2\right]}{0.1\mathrm{kg}}}$

$=0.283\mathrm{m}/\mathrm{s}$