Answer:
0.67 s
Explanation:
This is a simple harmonic motion (SHM).
The displacement, [tex]x[/tex], of an SHM is given by
[tex]x = A\cos(\omega t)[/tex]
A is the amplitude and [tex]\omega[/tex] is the angular frequency.
We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or [tex]\frac{\pi}{4}[/tex] radian.
From trigonometry, [tex]\sin A =\cos B[/tex] if A and B are complementary.
At [tex]t = 0[/tex], [tex]x = 3.5[/tex]
[tex]3.5 = A\cos(\omega \times0)[/tex]
[tex]A =3.5[/tex]
So
[tex]x = 3.5\cos(\omega t)[/tex]
At [tex]t = 0.12[/tex], [tex]x = 1.5[/tex]
[tex]1.5 = 3.5\cos(0.12\omega)[/tex]
[tex]\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286[/tex]
[tex]0.12\omega =\cos^{-1}0.4286[/tex]
[tex]0.12\omega = 1.13[/tex]
[tex]\omega = 9.4[/tex]
The period, [tex]T[/tex], is related to [tex]\omega[/tex] by
[tex]T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67[/tex]
