Answer and Explanation:
1. Evaluate the function x(t) at t=0.5
[tex]x(t)=1cos(2t+0.25)\\x(0.5)=cos(2*0.5+0.25)=cos(1+0.25)=cos(1.25)=0.3153223624\approx0.31cm[/tex]
2. The period of motion T can be calculated as:
[tex]T=\frac{2\pi}{\omega}[/tex]
Where:
[tex]\omega=2rad/s[/tex]
So:
[tex]T=\frac{2\pi}{2}=\pi\approx3.14s[/tex]
3. The angular frequency can be expressed as:
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
Solving for k:
[tex]k=(\omega)^2*m=(2)^2*4=4*4=16\frac{N}{m}[/tex]
4. Find the derivate of x(t):
[tex]\frac{dx}{dt} =v(t)=-2sin(2t+0.25)[/tex]
Now, the sine function reach its maximum value at π/2 so:
[tex]2t+025=\frac{\pi}{2}[/tex]
Solving for t:
[tex]t=\frac{\frac{\pi}{2} -0.25}{2} =0.6603981634s[/tex]
Evaluating v(t) for 0.6603981634:
[tex]v(0.6603981634)=-2sin(2*0.6603981634+0.25)=-2sin(\frac{\pi}{2} )=-2*1=2[/tex]
So the maximum speed of the block is:
[tex]v(0.6603981634)=2cm/s=0.02m/s[/tex]
In the negative direction of x-axis
5. The force is given by:
[tex]F=kx[/tex]
The cosine function reach its maximum value at 2π so:
[tex]2t+0.25=2\pi[/tex]
Solving for t:
[tex]t=\frac{2\pi-0.25}{2} =3.016592654s[/tex]
Evaluating x(t) for 3.016592654:
[tex]x(3.016592654)=cos(2*3.016592654+0.25)=cos(2\pi)=1cm=0.01m[/tex]
Therefore the the maximum force on the block is:
[tex]F=16*0.01=0.16N[/tex]
