Question

A 0.36 kg object, attached to a spring with constant k=10n/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude 0.082 m. What is it speed when it’s displacement is 0.041 m

Answer 1

Answer:

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

Explanation:

First, we need to determine the angular frequency of the system ($\omega$), in radians per second:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, in newtons per meter.

$m$ - Mass, in kilograms.

If we know that $k = 10\,\frac{N}{m}$ and $m = 0.36\,kg$, then the angular frequency of the system is:

$\omega = \sqrt{\frac{10\,\frac{N}{m} }{0.36\,kg} }$

$\omega \approx 5.270\,\frac{rad}{s}$

The kinematic formulas for the position ($x(t)$), in meters, velocity ($\dot x(t)$), in meters per second, and acceleration of the object ($\ddot x(t)$), in meters per square second, are:

$x(t) = A\cdot \cos \omega t$ (2)

$\dot x(t) = -\omega \cdot A \cdot \sin \omega t$ (3)

$\ddot x(t) = -\omega^{2}\cdot A \cdot \cos \omega t$ (4)

Where $A$ is the amplitude of the motion, in meters.

From (2) we determine the time associated with position $x(t) = 0.041\,m$ ($\omega \approx 5.270\,\frac{rad}{s}$, $A = 0.082\,m$):

$t = \frac{1}{\omega}\cdot \cos^{-1} \left(\frac{x(t)}{A} \right)$ (5)

$t = \frac{1}{5.270\,\frac{rad}{s} }\cdot \cos^{-1}\left(\frac{0.041\,m}{0.082\,m} \right)$

$t = 0.199\,s$

And the speed of the object is:

$\dot x(t) = -\left(5.270\,\frac{rad}{s} \right)\cdot (0.082\,m)\cdot \sin \left[\left(5.270\,\frac{rad}{s} \right)\cdot (0.199\,s)\right]$

$\dot x(t) \approx -0.375\,\frac{m}{s}$

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.