Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:
[tex]x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)[/tex]
for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:
[tex]\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}[/tex]
with this value you can compute the frequency:
a)
[tex]f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz[/tex]
b)
the mass of the block is given by the formula:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg[/tex]
c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:
[tex]\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s[/tex]
Finally, the amplitude is:
[tex]x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m[/tex]
