Question

an object moves in simple harmonic motion with amplitude 13 m and period 3 minutes. At time t= 0 minutes, its displacement d from rest is -13 m, and initially it moves in a positive direction.

Answer 1

Answer:

$x(t) = (13\,m)\cdot \cos \left[\frac{2\pi}{3}\cdot t \pm \pi\right]$, where t is measure in minutes.

Step-by-step explanation:

The statement consists in the construction of the motion function for a object experimenting a simple harmonic motion. The expression for simple harmonic motion is:

$x(t) = A\cdot \cos (\omega\cdot t + \phi)$

Where:

$A$ - Amplitude, in m.

$\omega$ - Angular frequency, in rad/s.

$\phi$ - Phase angle, in rad.

The angular frequency is:

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

$\omega = \frac{2\pi}{180\,s}$

$\omega = \frac{\pi}{90}$

The amplitude of the motion is 13 m and the phase angle is:

$(13\,m)\cdot \cos \phi = -13\,m$

$\cos \phi = -1$

$\phi = \pm\pi$

The position function for the object is:

$x(t) = (13\,m)\cdot \cos \left[\frac{2\pi}{3}\cdot t \pm \pi\right]$, where t is measure in minutes.