Answer:
a. vmax = 0.445 m/s
b. amx = 1.65 m/s^2
c. v = 0.32 m/s
d. 1.12 m/s^2
e. t = 0.21 s
Explanation:
a. The maximum speed of the object is given by the following formula:
[tex]v_{max}=\omega A[/tex] (1)
w: angular frequency of the object
A: amplitude of the motion = 12.2cm = 0.122m
The angular frequency is calculated by using:
[tex]\omega=\sqrt{\frac{k}{m}}[/tex] (2)
k: spring constant = 8.00 N/m
m: mass of the object = 0.600kg
you replace the equation (2) into the equation (1) and replace the values of the other parameters:
[tex]v_{max}=\sqrt{\frac{k}{m}}A=(\sqrt{\frac{8.00N/m}{0.600kg}})(0.122m)=0.445\frac{m}{s}[/tex]
The maximum speed is 0.445m/s
b. The maximum acceleration is:
[tex]a_{max}=\omega^2 A\\\omega=\sqrt{\frac{8.00N/m}{0.600kg}}=3.65\frac{rad}{s}\\\\a_{max}=(3.65rad/s)^2(0.122m)=1.625\frac{m}{s^2}[/tex]
The maximum acceleration is 1.65m/s^2
c. To calculate the value of the speed for x = 8.60cm you first find the time t by using the following equation of motion for a simple harmonic motion:
[tex]x=Asin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{x}{A})\\\\t=\frac{1}{3.65rad/s}sin^{-1}(\frac{8.60cm}{12.2cm})=0.21s[/tex]
You use this value of t in the following equation for v:
[tex]v=\omega Acos(\omega t)\\\\v=(3.65rad/s)(0.122m)cos((3.65rad/s)(0.21s))=0.32\frac{m}{s}[/tex]
The speed of the object when it is at x = 8.60cm is 0.32m/s
d. The acceleration is:
[tex]a=-\omega^2 A sin(\omega t)\\\\a=-(3.65rad/s)^2(0.122m)sin((3.65rad/s)(0.21))=1.12\frac{m}{s^2}[/tex]
The acceleration is 1.12 m/s^2
e. The time is 0.21s
