Answer:
(a) vmax = 0.34m/s
(b) v = 0.13m/s
(c) v = 0.31m/s
(d) x = 0.039m
Explanation:
Given information about the spring-mass system:
m: mass of the object = 0.30kg
k: spring constant = 22.6 N/m
A: amplitude of the motion = 4.0cm = 0.04m
(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 motion.
The angular frequency is calculated with the following relation:
[tex]\omega=\sqrt{\frac{k}{m}}[/tex] (2)
You replace the expression (2) into the equation (1) and replace the values of the parameters:
[tex]v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}[/tex]
The maximum speed of the object is 0.34 m/s
(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:
[tex]v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}[/tex]
The speed is 0.13m/s
(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:
[tex]x=Acos(\omega t)[/tex]
You solve the previous equation for t:
[tex]t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s[/tex]
With this value of t, you can calculate the speed of the object with the following formula:
[tex]v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}[/tex]
The speed of the object for x = 1.5cm is v = 0.31 m/s
(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:
[tex]\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s[/tex]
The position will be:
[tex]x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m[/tex]
The position of the object on which its speed is one-half its maximum velocity is 0.039
