(a) 4.06 cm
In a simple harmonic motion, the displacement is written as
[tex]x(t) = A cos (\omega t + \phi)[/tex] (1)
where
A is the amplitude
[tex]\omega[/tex] is the angular frequency
[tex]\phi[/tex] is the phase
t is the time
The displacement of the piston in the problem is given by
[tex]x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})[/tex] (2)
By putting t=0 in the formula, we find the position of the piston at t=0:
[tex]x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm[/tex]
(b) -14.69 cm/s
In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:
[tex]v(t) = x'(t) = -\omega A sin (\omega t + \phi)[/tex] (3)
Differentiating eq.(2), we find
[tex]v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})[/tex]
And substituting t=0, we find the velocity at time t=0:
[tex]v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s[/tex]
(c) -101.13 cm/s^2
In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:
[tex]a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)[/tex]
Differentiating eq.(3), we find
[tex]a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})[/tex]
And substituting t=0, we find the acceleration at time t=0:
[tex]a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2[/tex]
(d) 5.00 cm, 1.26 s
By comparing eq.(1) and (2), we notice immediately that the amplitude is
A = 5.00 cm
For the period, we have to start from the relationship between angular frequency and period T:
[tex]\omega=\frac{2\pi}{T}[/tex]
Using [tex]\omega = 5.0 rad/s[/tex] and solving for T, we find
[tex]T=\frac{2\pi}{5 rad/s}=1.26 s[/tex]
