Answer:
A) [tex]k=2.63*10^{5} N/m[/tex].
B)[tex]v=2.10m/s[/tex]
C)[tex]a=90.0m/s^{2}[/tex]
Explanation:
This problem is a simple harmonic motion problem. The equation of motion for the SHM is:
[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],
where x is the displacement of the mass about its point of equilibrium, t is time, and [tex]\omega[/tex] is the angular frequency.
A)
First, we need to remember that
[tex]\omega^{2}=\frac{k}{m}[/tex],
where k is the spring constant, and m is the mass.
From here we can simply solve for k, so
[tex]k=\omega^{2}m[/tex].
Now, we need to make use of an equation that relates the frequency and angular frequency. The equetion is
[tex]\omega=2\pi \nu[/tex],
where [tex]\nu[/tex] is the frequency. This leads us to
[tex]k=(2\pi \nu)^{2}m[/tex],
[tex]k=142(2*6.85*\pi)^{2}[/tex],
[tex]k=2.63*10^{5} N/m[/tex],
B) In simple harmonic motion, the velocity behaves as follow:
[tex]v=\omega Acos(\omega t)[/tex] (this is obtained by solving the equation of motion of the mass for the displacement x and take the derivative),
where A is the amplitude of the motion. Since we want the maximum value for the speed, we make [tex]cos(\omega t)=1[/tex] (this because cosine function goes from -1 to 1). With this, the maximum speed is simply
[tex]v = \omega A\\v=(2\pi \nu)A\\v=(2*6.85*\pi)*0.0486\\v=2.10m/s[/tex]
C) Here we are going to use the equation of motion of SHM
[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],
we know that
[tex]a=\frac{d^{2}x}{dt^{2}}[/tex] , where a is the acceleration,
[tex]a+\omega^{2}x=0\\a=-\omega^{2}x[/tex]
in this case, x goes from -A to A, so for a to be maximum we need that [tex]x=-A[/tex] ,and we get
[tex]a=-\omega^{2}(-A)\\a=\omega^{2}A\\a=(2\pi \nu)^{2}A\\a=(2*6.85*\pi)^{2}(0.0486)\\a=90.0m/s^{2}[/tex]
