a) [tex] x(t)=2.0 sin (10 t) [m][/tex]
The equation which gives the position of a simple harmonic oscillator is:
[tex]x(t)= A sin (\omega t)[/tex]
where
A is the amplitude
[tex]\omega=\sqrt{\frac{k}{m}}[/tex] is the angular frequency, with k being the spring constant and m the mass
t is the time
Let's start by calculating the angular frequency:
[tex]\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s[/tex]
The amplitude, A, can be found from the maximum velocity of the spring:
[tex]v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m[/tex]
So, the equation of motion is
[tex]x(t)= 2.0 sin (10 t) [m][/tex]
b) t=0.10 s, t=0.52 s
The potential energy is given by:
[tex]U(x)=\frac{1}{2}kx^2[/tex]
While the kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
The velocity as a function of time t is:
[tex]v(t)=v_{max} cos(\omega t)[/tex]
The problem asks as the time t at which U=3K, so we have:
[tex]\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}[/tex]
However, [tex]\frac{m}{k}=\frac{1}{\omega^2}[/tex], so we have
[tex](tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\[/tex]
with two solutions:
[tex]\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s[/tex]
[tex]\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s[/tex]
c) 3 seconds.
When x=0, the equation of motion is:
[tex]0=A sin (\omega t)[/tex]
so, t=0.
When x=1.00 m, the equation of motion is:
[tex]1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s[/tex]
So, the time needed is 3 seconds.
d) 0.097 m
The period of the oscillator in this problem is:
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s[/tex]
The period of a pendulum is:
[tex]T=2 \pi \sqrt{\frac{L}{g}}[/tex]
where L is the length of the pendulum. By using T=0.628 s, we find
[tex]L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m[/tex]
