Question

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20m/s to the right and an acceleration of 8.40m/s2 to the left.

How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Answer 1

Concepts and reason

The concept used to solve this problem is simple harmonic motion, Newton’s second law, restoring force, and conservation of energy.

Use the relation between the acceleration and object displacement to calculate the angular velocity.

Use the relation between the object displacement, velocity of the object, and angular velocity to calculate the amplitude of the wave.

Use the relation between object displacement and amplitude of the wave to calculate the distance moved by the object before it stops momentarily.

Fundamentals

Expression for the acceleration of the object in the simple harmonic is,

$a = - {\omega ^2}x$

Here, $a$ is the acceleration of the object, $\omega$ is the angular velocity, and x is the displacement of the object.

Rearrange the above equation to get the angular velocity,

$\omega = \sqrt {\frac{a}{x}}$

Expression for the velocity of the object is,

$v = \omega \sqrt {{A^2} - {x^2}}$

Here, v is the velocity of the object and A is the amplitude is the wave.

Rearrange the above equation to get the amplitude of the wave,

$A = \sqrt {{{\left( {\frac{v}{\omega }} \right)}^2} + {x^2}}$

Expression for the distance moved by the object before it stops momentarily is,

$X = A - x$

Here, X is the distance moved by the object before it stops momentarily.

Expression for the angular velocity is,

$\omega = \sqrt {\frac{a}{x}}$

Substitute $8.40\,{\rm{m/}}{{\rm{s}}^2}$ for $a$ and $0.600\,{\rm{m}}$ for x.

$\begin{array}{c}\\\omega = \sqrt {\frac{{8.40\,{\rm{m/}}{{\rm{s}}^2}}}{{0.600\,{\rm{m}}}}} \\\\ = 3.74\,{\rm{rad/s}}\\\end{array}$

Expression for the amplitude of the wave is,

$A = \sqrt {{{\left( {\frac{v}{\omega }} \right)}^2} + {x^2}}$

Substitute $2.20\,{\rm{m/s}}$ for v, $3.74\,{\rm{rad/s}}$for $\omega$, and $0.600\,{\rm{m}}$ for x.

$\begin{array}{c}\\A = \sqrt {{{\left( {\frac{{2.20\,{\rm{m/s}}}}{{3.74\,{\rm{rad/s}}}}} \right)}^2} + {{\left( {0.600\,{\rm{m}}} \right)}^2}} \\\\ = 0.840\,{\rm{m}}\\\end{array}$

Expression for the distance moved by the object before it stops momentarily is,

$X = A - x$

Substitute $0.840\,{\rm{m}}$ for A and $0.600\,{\rm{m}}$ for x.

$\begin{array}{c}\\X = 0.840\,{\rm{m}} - 0.600\,{\rm{m}}\\\\ = 0.240\,{\rm{m}}\\\end{array}$

Ans:

The object will move $0.240\,{\rm{m}}$before it stops momentarily and then starts to move back to the left.