Given data:
* The equation of the position is,
[tex]x=12\cos (6\pi t+\frac{\pi}{6})[/tex]Solution:
(1). From the given equation of the position, the amplitude of the displacement is,
[tex]x_o=12\text{ m}[/tex]Thus, the amplitude of the oscillation is 12 meter.
(2). The angular frequency of the oscillation from the given equation is,
[tex]\begin{gathered} \omega=6\pi\text{ } \\ \omega=18.85\text{ rad/s} \end{gathered}[/tex]Thus, the angular frequency of the oscillation is 18.85 radian per second.
(3). The linear frequency of the oscillation is,
[tex]\begin{gathered} f=\frac{2\pi}{\omega} \\ f=\frac{2\pi}{6\pi} \\ f=\frac{1}{3} \\ f=0.33\text{ Hz} \end{gathered}[/tex]Thus, the frequency of the oscillation is 0.33 Hz.
(4). The time period of oscillation is,
[tex]\begin{gathered} T=\frac{1}{f} \\ T=\frac{1}{0.33} \\ T=3\text{ s} \end{gathered}[/tex]Thus, the time period of the oscillation is 3 seconds.
(5). The position of the oscillator at tiem t = 0 s is,
[tex]\begin{gathered} x=12\cos (6\pi\times0+\frac{\pi}{6}) \\ x=12\text{cos(}\frac{\pi}{6}\text{)} \\ x=11.99 \\ x\approx12\text{ m} \end{gathered}[/tex]Thus, the position of the oscillator at time t=0 seconds is 12 meter.
(6). The maximum speed of the oscillator is,
[tex]\begin{gathered} v=x_o\omega \\ v=12\times18.85 \\ v=226.2\text{ m/s} \end{gathered}[/tex]Thus, the maximum value of velocity is 226.2 m/s.
(7). The maximum acceleration of the oscillator is,
[tex]\begin{gathered} a=\omega^2x_o \\ a=(18.85)^2\times12 \\ a=4263.9ms^{-2} \end{gathered}[/tex]Thus, the maximum acceleration of the oscillator is 4263.9 meter per second squared.
(8). The displacement of the oscillator at the 5 second is,
[tex]\begin{gathered} x=12\cos (6\pi t+\frac{\pi}{6}) \\ x=12\cos (6\pi\times5+\frac{\pi}{6}) \\ x=-0.998 \\ x\approx-1\text{ m} \end{gathered}[/tex]Thus, the displacement of oscillator after 5 seconds is -1 meter.
(9). The velocity of the oscillator in terms of time is,
[tex]\begin{gathered} v=\frac{dx}{dt} \\ v=\frac{d}{dt}(12\cos (6\pi t+\frac{\pi}{6})) \\ v=12\times6\pi\times-\sin (6\pi t+\frac{\pi}{6}) \\ v=-72\pi\sin (6\pi t+\frac{\pi}{6}) \end{gathered}[/tex]The velocity of the oscillator at time 5 seconds is,
[tex]\begin{gathered} v=-72\pi\sin (6\pi\times5+\frac{\pi}{6}) \\ v=-225.4\text{ m/s} \end{gathered}[/tex]Thus, the velocity of the oscillator at 5 seconds is -225.4 meter per second.
(10). The acceleration of the oscillator is,
[tex]\begin{gathered} a=\frac{dv}{dt} \\ a=\frac{d}{dt}(-72\pi\sin (6\pi t+\frac{6}{\pi})) \\ a=-72\pi\times6\pi\cos (6\pi t+\frac{6}{\pi}) \end{gathered}[/tex]The value of acceeration at tiem 5 seconds is,
[tex]\begin{gathered} a=-72\pi\times6\pi\cos (6\pi\times5+\frac{6}{\pi}) \\ a=354.65ms^{-2} \end{gathered}[/tex]Thus, the acceleration of the oscillator at 5 second is 354.65 meter per second squared.
