Answer:
The amplitude and period are 10.1 cm and 0.4 s.
Explanation:
Given that,
The maximum acceleration of an object in SHM, [tex]a=8\pi\ m/s^2[/tex]
The maximum velocity of the object, [tex]v=1.6\ m/s[/tex]
The formula for the maximum velocity and acceleration in SHM is given by :
[tex]v=A\omega\ ....(1)\\\\a=A\omega^2\ ....(2)[/tex]
Dividing equation (1) and (2) :
[tex]\dfrac{v}{a}=\dfrac{A\omega}{A\omega^2}\\\\\dfrac{v}{a}=\dfrac{1}{\omega}\\\\\omega=\dfrac{a}{v}\\\\\dfrac{2\pi}{T}=\dfrac{a}{v}\\\\T=\dfrac{2\pi v}{a}\\\\T=\dfrac{2\pi \times 1.6}{8\times \pi}\\\\T=0.4\ s[/tex]
Also,
[tex]\omega=\dfrac{8\pi}{1.6}\\\\=15.7\ rad/s[/tex]
Pu in equation (1) to find A,
[tex]A=\dfrac{v}{\omega}\\\\=\dfrac{1.6}{15.7}\\\\=0.101\ m\\\\=10.1\ cm[/tex]
So, the amplitude and period is 10.1 cm and 0.4 s.
